Coordinate Geometry - Parabola

Find the vertex, axis, focus and directrix of the parabola $\;$ $\left(x - 4\right)^2 = -8y + 5$.

Equation of given parabola: $\;$ $\left(x - 4\right)^2 = -8y + 5$

i.e. $\;$ $\left(x - 4\right)^2 = -8 \left(y - \dfrac{5}{8}\right)$ $\;\;\; \cdots \; (1)$

To shift the origin to the point $\left(4, \dfrac{5}{8}\right)$,

let $\;$ $x - 4 = X$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $y - \dfrac{5}{8} = Y$ $\;\;\; \cdots \; (2b)$

In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes

$X^2 = -8Y$ $\;\;\; \cdots \; (3)$

1. Vertex of equation $(3)$ is $\;$ $\left(0, 0\right)$
   
   i.e. $\;$ $X = 0, \;\;\; Y = 0$
   
   When $\;$ $X = 0$
   
   $\implies$ $x - 4 = 0$ $\;$ i.e. $\;$ $x = 4$ $\;\;\;$ [by equation $(2a)$]
   
   When $\;$ $Y = 0$
   
   $\implies$ $y - \dfrac{5}{8} = 0$ $\;$ i.e. $\;$ $y = \dfrac{5}{8}$ $\;\;\;$ [by equation $(2b)$]
   
   $\therefore \;$ Vertex of equation $(1)$ is $\;$ $\left(4, \dfrac{5}{8}\right)$
   
   
2. Axis of equation $(3)$ is $\;\;$ $X = 0$
   
   $\therefore \;$ Axis of equation $(1)$ is $\;\;\;$ $x - 4 = 0$ $\;\;\;$ [by equation $(2a)$]
   
   i.e. $\;$ $x = 4$
   
   
3. Comparing equation $(3)$ with the standard equation of parabola $\;\;$ $X^2 = -4a Y$ $\;\;$ gives
   
   $4a = 8$ $\implies$ $a = 2$
   
   $\therefore \;$ Focus of equation $(3)$ is $\;\;\;$ $\left(0, -a\right) = \left(0, -2\right)$
   
   $\implies$ $X = 0, \;\;\; Y = -2$
   
   When $\;$ $X = 0$
   
   $\implies$ $x - 4 = 0$ $\;$ i.e. $\;$ $x = 4$ $\;\;\;$ [by equation $(2a)$]
   
   When $\;$ $Y = -2$
   
   $\implies$ $y - \dfrac{5}{8} = -2$ $\;$ i.e. $\;$ $y = \dfrac{-11}{8}$ $\;\;\;$ [by equation $(2b)$]
   
   $\therefore \;$ Focus of equation $(1)$ is $\;\;\;$ $\left(4, \dfrac{-11}{8}\right)$
   
   
4. Directrix of equation $(3)$ is $\;\;\;$ $Y = a$
   
   i.e. $\;$ $Y = 2$
   
   i.e. $y - \dfrac{5}{8} = 2$ $\;\;\;$ [by equation $(2b)$]
   
   i.e. $\;$ $y= \dfrac{21}{8}$
   
   $\therefore \;$ Directrix of equation $(1)$ is $\;\;\;$ $8y = 21$

Coordinate Geometry - Parabola

Find the vertex, axis, focus and directrix of the parabola $\;$ $\left(y + 3\right)^2 = 4x + 4$.

Equation of given parabola: $\;$ $\left(y + 3\right)^2 = 4x + 4$

i.e. $\;$ $\left(y + 3\right)^2 = 4 \left(x + 1\right)$ $\;\;\; \cdots \; (1)$

To shift the origin to the point $\left(-1, -3\right)$,

let $\;$ $x + 1 = X$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $y + 3 = Y$ $\;\;\; \cdots \; (2b)$

In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes

$Y^2 = 4 X$ $\;\;\; \cdots \; (3)$

1. Vertex of equation $(3)$ is $\;$ $\left(0, 0\right)$
   
   i.e. $\;$ $X = 0, \;\;\; Y = 0$
   
   When $\;$ $X = 0$
   
   $\implies$ $x + 1 = 0$ $\;$ i.e. $\;$ $x = -1$ $\;\;\;$ [by equation $(2a)$]
   
   When $\;$ $Y = 0$
   
   $\implies$ $y + 3 = 0$ $\;$ i.e. $\;$ $y = -3$ $\;\;\;$ [by equation $(2b)$]
   
   $\therefore \;$ Vertex of equation $(1)$ is $\;$ $\left(-1, -3\right)$
   
   
2. Axis of equation $(3)$ is $\;\;$ $Y = 0$
   
   $\therefore \;$ Axis of equation $(1)$ is $\;\;\;$ $y + 3 = 0$ $\;\;\;$ [by equation $(2b)$]
   
   i.e. $\;$ $y = -3$
   
   
3. Comparing equation $(3)$ with the standard equation of parabola $\;\;$ $Y^2 = 4a X$ $\;\;$ gives
   
   $a = 1$
   
   $\therefore \;$ Focus of equation $(3)$ is $\;\;\;$ $\left(a, 0\right) = \left(1, 0\right)$
   
   $\implies$ $X = 1, \;\;\; Y = 0$
   
   When $\;$ $X = 1$
   
   $\implies$ $x + 1 = 1$ $\;$ i.e. $\;$ $x = 0$ $\;\;\;$ [by equation $(2a)$]
   
   When $\;$ $Y = 0$
   
   $\implies$ $y + 3 = 0$ $\;$ i.e. $\;$ $y = -3$ $\;\;\;$ [by equation $(2b)$]
   
   $\therefore \;$ Focus of equation $(1)$ is $\;\;\;$ $\left(0, -3\right)$
   
   
4. Directrix of equation $(3)$ is $\;\;\;$ $X = -a$
   
   i.e. $\;$ $X = -1$
   
   i.e. $x + 1 = -1$ $\;\;\;$ [by equation $(2a)$]
   
   $\therefore \;$ Directrix of equation $(1)$ is $\;\;\;$ $x + 2 = 0$

Coordinate Geometry - Circle

Find the equation of the circle whose center is $\left(5, \dfrac{\pi}{3}\right)$ and radius $7$.

Given: $\;\;$ Center of the circle $= \left(R, \phi\right) = \left(5, \dfrac{\pi}{3}\right)$

i.e. $\;$ $R = 5, \;\; \phi = \dfrac{\pi}{3}$

Radius of the circle $= a = 7$

Let the equation of the required circle be

$R^2 + r^2 - 2 R r \cos \left(\theta - \phi\right) = a^2$

Substituting the values of $R$, $\phi$ and $a$, we get

$25 + r^2 - 2 \times 5 \times r \cos \left(\theta - \dfrac{\pi}{3}\right) = 49$

i.e. $\;$ $r^2 - 10r \left(\cos \theta \cos \dfrac{\pi}{3} + \sin \theta \sin \dfrac{\pi}{3}\right) - 24 = 0$

i.e. $\;$ $r^2 - 10r \left(\dfrac{1}{2} \cos \theta + \dfrac{\sqrt{3}}{2} \sin \theta\right) - 24 = 0$

i.e. $\;$ $r^2 - 5r \left(\cos \theta + \sqrt{3} \sin \theta\right) - 24 = 0$

Coordinate Geometry - Circle

Find the center and radius of the circle $\;$ $r^2 - 2r \left(24 \cos \theta - 7 \sin \theta\right) + 141 = 0$

Equation of given circle: $\;$ $r^2 - 2r \left(24 \cos \theta - 7 \sin \theta\right) + 141 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $\left(R, \phi\right)$ $\;$ be the polar coordinates of the center and the radius be $= a$.

Then equation $(1)$ should be identical with

$r^2 - 2R r \cos \left(\theta - \phi\right) + R^2 - a^2 = 0$

i.e. $\;$ $r^2 - 2R r \left(\cos \theta \cos \phi + \sin \theta \sin \phi\right) + R^2 - a^2 = 0$ $\;\;\; \cdots \; (2)$

$\therefore \;$ $-4r \cos \theta = -2 R r \cos \theta \cos \phi$

i.e. $\;$ $R \cos \phi = 24$ $\;\;\; \cdots \; (3a)$;

$14 r \sin \theta = -2 R r \sin \theta \sin \phi$

i.e. $\;$ $R \sin \phi = -7$ $\;\;\; \cdots \; (3b)$

and $\;$ $R^2 - a^2 = 141$ $\;\;\; \cdots \; (3c)$

From equations $(3a)$ and $(3b)$

$\tan \phi = \dfrac{-7}{24}$

$\implies$ $\phi = \tan^{-1} \left(\dfrac{-7}{24}\right)$ $\;\;\; \cdots \; (4a)$

and $\;$ $R^2 \cos^2 \phi = 576, \;\; R^2 \sin^2 \phi = 49$

i.e. $\;$ $R^2 \left(\sin^2 \phi + \cos^2 \phi\right) = R^2 = 579 + 46 = 625$

$\implies$ $R = 25$

Substituting the value of $R^2$ in equation $(3c)$ gives

$625 - a^2 = 141$

i.e. $\;$ $a^2 = 484$ $\implies$ $a = 22$ $\;\;\; \cdots \; (4b)$

$\therefore \;$ The center of the circle is $\;$ $\left(25, \tan^{-1} \left(\dfrac{-7}{24}\right)\right)$ $\;$ and the radius is $\;$ $22$.

Coordinate Geometry - Circle

Find the center and radius of the circle $\;$ $r^2 - 4r \left(\sqrt{3} \cos \theta + \sin \theta\right) + 7 = 0$

Equation of given circle: $\;$ $r^2 - 4r \left(\sqrt{3} \cos \theta + \sin \theta\right) + 7 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $\left(R, \phi\right)$ $\;$ be the polar coordinates of the center and the radius be $= a$.

Then equation $(1)$ should be identical with

$r^2 - 2R r \cos \left(\theta - \phi\right) + R^2 - a^2 = 0$

i.e. $\;$ $r^2 - 2R r \left(\cos \theta \cos \phi + \sin \theta \sin \phi\right) + R^2 - a^2 = 0$

$\therefore \;$ $4r \times \sqrt{3} \cos \theta = 2 R r \cos \theta \cos \phi$

i.e. $\;$ $R \cos \phi = 2 \sqrt{3}$ $\;\;\; \cdots \; (3a)$;

$4 r \sin \theta = 2 R r \sin \theta \sin \phi$

i.e. $\;$ $R \sin \phi = 2$ $\;\;\; \cdots \; (3b)$

and $\;$ $R^2 - a^2 = 7$ $\;\;\; \cdots \; (3c)$

From equations $(3a)$ and $(3b)$

$\tan \phi = \dfrac{1}{\sqrt{3}}$

$\implies$ $\phi = \tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\pi}{6}$ $\;\;\; \cdots \; (4a)$

and $\;$ $R^2 \cos^2 \phi = 12, \;\; R^2 \sin^2 \phi = 4$

i.e. $\;$ $R^2 \left(\sin^2 \phi + \cos^2 \phi\right) = R^2 = 12 + 4 = 16$

$\implies$ $R = 4$

Substituting the value of $R^2$ in equation $(3c)$ gives

$16 - a^2 = 7$

i.e. $\;$ $a^2 = 9$ $\implies$ $a = 3$ $\;\;\; \cdots \; (4b)$

$\therefore \;$ The center of the circle is $\;$ $\left(4, \dfrac{\pi}{6}\right)$ $\;$ and the radius is $\;$ $3$.

Coordinate Geometry - Circle

Determine the limiting points of the system of circles $\;$ $x^2 + y^2 -3x - 9y + 45 - \lambda \left(x^2 + y^2 -2x -6y + 30\right) = 0$

Given equation of system of circles

$x^2 + y^2 -3x - 9y + 45 - \lambda \left(x^2 + y^2 -2x -6y + 30\right) = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ can be written as

$x^2 \left(1 - \lambda\right) + y^2 \left(1 - \lambda\right) + \left(2 \lambda - 3\right) x + \left(6 \lambda - 9\right) y + \left(45 - 30 \lambda\right) = 0$

i.e. $\;$ $x^2 + y^2 + \left(\dfrac{2 \lambda - 3}{1 - \lambda}\right) x + \left(\dfrac{6 \lambda - 9}{1 - \lambda}\right) y + \dfrac{45 - 30 \lambda}{1 - \lambda} = 0$ $\;\;\; \cdots \; (2)$

Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$g = \dfrac{2 \lambda - 3}{2 - 2 \lambda}, \;\; f = \dfrac{6 \lambda - 9}{2 - 2 \lambda}, \;\; c = \dfrac{45 - 30 \lambda}{1 - \lambda}$

Center of equation $(2)$ $= \left(-g, -f\right) = \left(\dfrac{3 - 2 \lambda}{2 - 2 \lambda}, \dfrac{9 - 6 \lambda}{2 - 2\lambda}\right)$ $\;\;\; \cdots \; (3a)$

Radius of equation $(2) = r = \sqrt{g^2 + f^2 - c}$

i.e. $\;$ $r = \sqrt{\left(\dfrac{2 \lambda - 3}{2 - 2 \lambda}\right)^2 + \left(\dfrac{6 \lambda - 9}{2 - 2 \lambda}\right)^2 - \left(\dfrac{45 - 30 \lambda}{1 - \lambda}\right)}$ $\;\;\; \cdots \; (3b)$

The limiting points are the centers of circles of the system whose radii are zero.

$\therefore \;$ We have from equation $(3b)$

$\left(\dfrac{2 \lambda - 3}{2 - 2 \lambda}\right)^2 + \left(\dfrac{6 \lambda - 9}{2 - 2 \lambda}\right)^2 - \left(\dfrac{45 - 30 \lambda}{1 - \lambda}\right) = 0$

i.e. $\;$ $\dfrac{4 \lambda^2 - 12 \lambda + 9 + 36 \lambda^2 - 108 \lambda + 81}{4 \left(1 - \lambda\right)^2} = \dfrac{15 \left(3 - 2 \lambda\right)}{1 - \lambda}$

i.e. $\;$ $\dfrac{40 \lambda^2 - 120 \lambda + 90}{4 \left(1 - \lambda\right)} = 15 \left(3 - 2 \lambda\right)$ $\;\;\;$ provided $\;$ $1 - \lambda \neq 0$

i.e. $\;$ $20 \lambda^2 - 60 \lambda + 45 = 30 \left(3 - 2 \lambda\right) \left(1 - \lambda\right)$

i.e. $\;$ $4 \lambda^2 - 12 \lambda + 9 = 6 \left(3 - 5 \lambda + 12 \lambda^2\right)$

i.e. $\;$ $8 \lambda^2 - 18 \lambda + 9 = 0$

$\implies$ $\lambda = \dfrac{3}{4}$ $\;$ or $\;$ $\lambda = \dfrac{3}{2}$

Substituting the values of $\lambda$ in equation $(3a)$ gives the limiting points.

When $\;$ $\lambda = \dfrac{3}{4}$, $\;$ the limiting point is

$\left(\dfrac{3 - 2 \times \dfrac{3}{4}}{2 - 2 \times\dfrac{3}{4}}, \; \dfrac{9 - 6 \times \dfrac{3}{4}}{2 - 2 \times \dfrac{3}{4}}\right) = \left(3, 9\right)$

When $\;$ $\lambda = \dfrac{3}{2}$, $\;$ the limiting point is

$\left(\dfrac{3 - 2 \times \dfrac{3}{2}}{2 - 2 \times \dfrac{3}{2}}, \; \dfrac{9 - 6 \times \dfrac{3}{2}}{2 - 2 \times \dfrac{3}{2}}\right) = \left(0, 0\right)$

Coordinate Geometry - Circle

Find the common tangents of the circles $\;$ $x^2 + y^2 - 3x - 4y = 0$, $\;$ $x^2 + y^2 - 21x + 90 = 0$.

The given circles are

$x^2 + y^2 - 3x - 4y = 0$ $\;\;\; \cdots \; (1)$

$x^2 + y^2 - 21x + 90 = 0$ $\;\;\; \cdots \; (2)$

Comparing equations $(1)$ and $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

for circle $(1)$: $\;$ $g_1 = \dfrac{-3}{2}, \;\; f_1 = -2, \;\; c_1 = 0$

center of circle $= C_1 = \left(-g_1, -f_1\right) = \left(\dfrac{3}{2}, 2\right)$,

radius of circle $= r_1 = \sqrt{g_1^2 + f_1^2 - c_1} = \sqrt{\dfrac{9}{4} + 4 - 0} = \dfrac{5}{2}$

for circle $(2)$: $\;$ $g_2 = \dfrac{-21}{2}, \;\; f_2 = 0, \;\; c_2 = 90$

center of circle $= C_2 = \left(-g_2, -f_2\right) = \left(\dfrac{21}{2}, 0\right)$

radius of circle $= r_2 = \sqrt{g_2^2 + f_2^2 - c_2} = \sqrt{\dfrac{441}{4} + 0 - 90} = \dfrac{9}{2}$

Ratio $\;$ $\dfrac{r_1}{r_2} = \dfrac{5}{9}$

Let $S_1$ be the external center of similitude of circles $(1)$ and $(2)$.

Then, the point $S_1$ divides the line joining $C_1C_2$ externally in the ratio $\dfrac{r_1}{r_2}$.

$\therefore \;$ $S_1 = \left(\dfrac{5 \times \dfrac{21}{2} - 9 \times \dfrac{3}{2}}{5-9}, \dfrac{5 \times 0 - 9 \times 2}{5-9}\right) = \left(\dfrac{-39}{4}, \dfrac{9}{2}\right)$

Let $S_2$ be the internal center of similitude of circles $(1)$ and $(2)$.

Then, the point $S_2$ divides the line joining $C_1C_2$ internally in the ratio $\dfrac{r_1}{r_2}$.

$\therefore \;$ $S_2 = \left(\dfrac{5 \times \dfrac{21}{2} + 9 \times \dfrac{3}{2}}{5+9}, \dfrac{5 \times 0 + 9 \times 2}{5+9}\right) = \left(\dfrac{33}{7}, \dfrac{9}{7}\right)$

Let the equation to either of the tangents passing through $S_1$ be

$y - \dfrac{9}{2} = m \left(x + \dfrac{39}{4}\right)$

i.e. $\;$ $mx - y + \dfrac{39}{4}m + \dfrac{9}{2} = 0$ $\;\;\; \cdots \; (3)$

Equation $(3)$ will touch circle $(1)$ if perpendicular distance from $C_1$ to equation $(3)$ is equal to $r_1$.

i.e. $\;$ $\dfrac{\dfrac{3}{2}m - 2 + \dfrac{39}{4}m + \dfrac{9}{2}}{\sqrt{m^2 + 1}} = \pm \dfrac{5}{2}$

i.e. $\;$ $\dfrac{45m + 10}{4 \sqrt{m^2 + 1}} = \pm \dfrac{5}{2}$

i.e. $\;$ $45m + 10 = \pm 10 \sqrt{m^2 + 1}$

i.e. $\;$ $2025 m^2 + 900 m + 100 = 100 + 100 m^2$

i.e. $\;$ $1925 m^2 + 900 m = 0$ $\;\;\; \cdots \; (4)$

Solving equation $(4)$ we get $\;\;$ $m = 0$, $\;$ or $\;$ $m = \dfrac{-36}{77}$

Substituting the value of $m$ in equation $(3)$ gives the required tangents.

$\therefore \;$ The required tangents are

$-y + \dfrac{9}{2} = 0$; $\;\;\;$ $\dfrac{-36}{77}x - y + \dfrac{39}{4} \times \left(\dfrac{-36}{77}\right) + \dfrac{9}{2} = 0$

i.e. $\;$ $2y - 9 = 0$ $\;\;\; \cdots \; (5a)$; $\;\;\;$ $36 x + 77y + \dfrac{9}{2} = 0$ $\;\;\; \cdots \; (5b)$

Let the equation to either of the tangents passing through $S_2$ be

$y - \dfrac{9}{7} = m_1 \left(x - \dfrac{33}{7}\right)$

i.e. $\;$ $m_1 x - y + \dfrac{9}{7} - \dfrac{33}{7} m = 0$ $\;\;\; \cdots \; (6)$

Equation $(6)$ will touch circle $(1)$ if perpendicular distance from $C_1$ to equation $(6)$ is equal to $r_1$.

i.e. $\;$ $\dfrac{\dfrac{3}{2} m_1 - 2 + \dfrac{9}{7} - \dfrac{33}{7} m_1}{\sqrt{m_1^2 + 1}} = \pm \dfrac{5}{2}$

i.e. $\;$ $\dfrac{-45 m_1 - 10}{14\sqrt{m_1^2 + 1}} = \pm \dfrac{5}{2}$

i.e. $\;$ $-45m_1 - 10 = \pm 35 \sqrt{m_1^2 + 1}$

i.e. $\;$ $2025 m_1^2 + 900 m_1 + 100 = 1225 + 1225 m_1^2$

i.e. $\;$ $32m_1^2 + 36m_1 - 45 = 0$ $\;\;\; \cdots \; (7)$

Solving equation $(7)$ we get $\;\;$ $m_1 = \dfrac{-15}{8}$, $\;$ or $\;$ $m_1 = \dfrac{3}{4}$

Substituting the value of $m_1$ in equation $(6)$ gives the required tangents.

$\therefore \;$ The required tangents are

$\dfrac{-15}{8} x - y + \dfrac{9}{7} - \dfrac{33}{7} \times \left(\dfrac{-15}{8}\right) = 0$; $\;\;\;$ $\dfrac{3}{4} x - y + \dfrac{9}{7} - \dfrac{33}{7} \times \dfrac{3}{4} = 0$

i.e. $\;$ $15x + 8y - 81 = 0$ $\;\;\; \cdots \; (8a)$; $\;\;\;$ $3x - 4y - 9 = 0$ $\;\;\; \cdots \; (8b)$

Equations $(5a)$, $(5b)$, $(8a)$ and $(8b)$ are the required common tangents to circles given by equations $(1)$ and $(2)$.

Coordinate Geometry - Circle

Find the equation of the circle which passes through the points of intersection of $\;$ $x^2 + y^2 + 4x - 6y - 12 = 0$ $\;$ and $\;$ $x^2 + y^2 -5x + 7y - 19 = 0$ $\;$ and has its center on $\;$ $x + y = 0$.

The given circles are

$x^2 + y^2 + 4x - 6y - 12 = 0$ $\;\;\; \cdots \; (1)$

$x^2 + y^2 - 5x + 7y - 19 = 0$ $\;\;\; \cdots \; (2)$

Equation of circle through the intersection of equations $(1)$ and $(2)$ is

$\left(x^2 + y^2 + 4x - 6y - 12\right) + k \left(x^2 + y^2 - 5x + 7y - 19\right) = 0$ $\;$ where $k$ is a constant

i.e. $\;$ $\left(1 + k\right) x^2 + \left(1 + k\right) y^2 + \left(4 - 5k\right) x + \left(7k - 6\right) y - 12 - 19k = 0$

i.e. $\;$ $x^2 + y^2 + \left(\dfrac{4 - 5k}{1 + k}\right) x + \left(\dfrac{7k - 6}{1 + k}\right) y + \left(\dfrac{-12 - 19k}{1 + k}\right) = 0$ $\;\;\; \cdots \; (3)$

Comparing equation $(3)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$g = \dfrac{4 - 5k}{2 + 2k}, \;\;\; f = \dfrac{7k - 6}{2 + 2k}, \;\;\; c = \dfrac{-12 - 19k}{1 + k}$

Center of the required circle $= \left(-g, -f\right) = \left(\dfrac{5k - 4}{2 + 2k}, \dfrac{6 - 7k}{2 + 2k}\right)$

Center lies on $\;$ $x + y = 0$ $\;\;\; \cdots \; (4)$

$\therefore \;$ We have,

$\dfrac{5k - 4}{2 + 2k} + \dfrac{6 - 7k}{2 + 2k} = 0$

i.e. $\;$ $5k - 4 + 6 - 7k = 0$

i.e. $\;$ $2k = 2$ $\implies$ $k = 1$

Substituting the value of $k$ in equation $(3)$ gives the equation of required circle as

$x^2 + y^2 + \left(\dfrac{4 - 5}{1 + 1}\right)x + \left(\dfrac{7 - 6}{1 + 1}\right)y + \left(\dfrac{-12 - 19}{1 + 1}\right) = 0$

i.e. $\;$ $2x^2 + 2y^2 -x + y - 31 = 0$

Coordinate Geometry - Circle

Find the equation to the circle whose diameter is the intercept made on $\;$ $3x + 4y = 1$ $\;$ by $\;$ $5x^2 + 6xy + y^2 = 0$.

Equation of given line: $\;$ $3x + 4y = 1$ $\;\;\; \cdots \; (1)$

Equation of given pair of lines: $\;$ $5x^2 + 6 xy + y^2 = 0$ $\;\;\; \cdots \; (2)$

i.e. $\;$ $5x^2 + 5xy + xy + y^2 = 0$

i.e. $\;$ $5x \left(x + y\right) + y \left(x + y\right) = 0$

i.e. $\;$ $\left(5x + y\right) \left(x + y\right) = 0$

i.e. $\;$ Equation $(2)$ represents the lines

$5x + y = 0$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $x + y = 0$ $\;\;\; \cdots \; (2b)$

Solving equations $(1)$ and $(2a)$ simultaneously gives the point $\;$ $P \left(x_1, y_1\right) = \left(\dfrac{-1}{17}, \dfrac{5}{17}\right)$

Solving equations $(1)$ and $(2b)$ simultaneously gives the point $\;$ $Q \left(x_2, y_2\right) = \left(-1, 1\right)$

As per the problem, points $P$ and $Q$ are the extremities of the diameter of the required circle.

$\therefore \;$ Equation of required circle is

$\left(x - x_1\right) \left(x - x_2\right) + \left(y - y_1\right) \left(y - y_2\right) = 0$

i.e. $\;$ $\left(x + \dfrac{1}{17}\right) \left(x + 1\right) + \left(y - \dfrac{5}{17}\right) \left(y - 1\right) = 0$

i.e. $\;$ $x^2 + \dfrac{18}{17} x + \dfrac{1}{17} + y^2 - \dfrac{22}{17} y + \dfrac{5}{17} = 0$

i.e. $\;$ $17 x^2 + 17 y^2 + 18 x - 22 y + 6 = 0$

Coordinate Geometry - Circle

Find the equation of the circle which passes through the points of intersection of $\;$ $x^2 + y^2 - 4x - 6y - 21 = 0$ $\;$ and $\;$ $3x + 4y + 5 = 0$ $\;$ and the point $\;$ $\left(1, 2\right)$.

Given circle: $\;\;$ $x^2 + y^2 - 4x - 6y - 21 = 0$ $\;\;\; \cdots \; (1)$

Given line: $\;\;$ $3x + 4y + 5 = 0$ $\;\;\; \cdots \; (2)$

Equation of circle passing through the intersection of equations $(1)$ and $(2)$ is

$\left(x^2 + y^2 - 4x - 6y - 21\right) + \lambda \left(3x + 4y + 5\right) = 0$ $\;\;$ where $\;$ $\lambda$ $\;$ is a constant.

i.e. $\;$ $x^2 + y^2 + \left(3 \lambda - 4\right) x + \left(4 \lambda - 6\right) y + 5 \lambda - 21 = 0$ $\;\;\; \cdots \; (3)$

Given: Equation $(3)$ passes through the point $\;$ $\left(1, 2\right)$

$\therefore \;$ We have

$1^2 + 2^2 + \left(3 \lambda - 4\right) \times 1 + \left(4 \lambda - 6\right) \times 2 + 5 \lambda - 21 = 0$

i.e. $\;$ $5 + 3 \lambda - 4 + 8 \lambda - 12 + 5 \lambda - 21 = 0$

i.e. $\;$ $16 \lambda - 32 = 0$ $\implies$ $\lambda = 2$ $\;\;\; \cdots \; (4)$

In view of equation $(4)$, equation $(3)$ becomes

$x^2 + y^2 + 2x + 2y - 11 = 0$

which is the required equation of circle.

Coordinate Geometry - Circle

Find the equation of the circle which cuts $\;$ $x^2 + y^2 + 4x + 7 = 0$, $\;$ $2x^2 + 2y^2 + 3x + 5y + 9 = 0$ $\;$ and $\;$ $x^2 + y^2 + y = 0$ $\;$ orthogonally.

Given circles:

$x^2 + y^2 + 4x + 7 = 0$ $\;\;\; \cdots \; (1)$

$2x^2 + 2y^2 + 3x + 5y + 9 = 0$

i.e. $\;$ $x^2 + y^2 + \dfrac{3}{2} x + \dfrac{5}{2} y + \dfrac{9}{2} = 0$ $\;\;\; \cdots \; (2)$

$x^2 + y^2 +y = 0$ $\;\;\; \cdots \; (3)$

Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ $\;$ gives

$g_1 = 2, \; f_1 = 0, \; c_1 = 7$

Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ $\;$ gives

$g_2 = \dfrac{3}{4}, \; f_2 = \dfrac{5}{4}, \; c_2 = \dfrac{9}{2}$

Comparing equation $(3)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_3 x + 2f_3 y + c_3 = 0$ $\;$ gives

$g_3 = 0, \; f_3 = \dfrac{1}{2}, \; c_3 = 0$

The radical axis of circles $(1)$ and $(2)$ is

$2 x \left(g_1 - g_2\right) + 2y \left(f_1 - f_2\right) + c_1 - c_2 = 0$

i.e. $\;$ $2x \left(2 - \dfrac{3}{4}\right) + 2y \left(0 - \dfrac{5}{4}\right) + 7 - \dfrac{9}{2} = 0$

i.e. $\;$ $\dfrac{5}{2} x - \dfrac{5}{2} y + \dfrac{5}{2} = 0$

i.e. $\;$ $x - y + 1 - 0$ $\;\;\; \cdots \; (4)$

The radical axis of circles $(1)$ and $(3)$ is

$2 x \left(g_1 - g_3\right) + 2y \left(f_1 - f_3\right) + c_1 - c_3 = 0$

i.e. $\;$ $2x \times \left(2 - 0\right) + 2y \left(0 - \dfrac{1}{2}\right) + 7 = 0$

i.e. $\;$ $4x - y + 7 = 0$ $\;\;\; \cdots \; (5)$

Solving equations $(4)$ and $(5)$ simultaneously gives

$x = -2, \; y = -1$

$\therefore \;$ The radical center is $\;$ $\left(h, k\right) = \left(-2, -1\right)$

Square of tangent from $\left(h, k\right)$ to equation $(1)$ is

$h^2 + k^2 + 2 g_1 h + 2 f_1 k + c_1$

$= 4 + 1 + 2 \times 2 \times \left(-2\right) + 2 \times 0 \times \left(-1\right) + 7 = 4$

$\therefore \;$ Required equation of circle is

$\left(x + 2\right)^2 + \left(y + 1\right)^2 = 4$

i.e. $\;$ $x^2 + 4x + 4 + y^2 + 2y + 1 = 4$

i.e. $\;$ $x^2 + y^2 + 4x + 2y + 1 = 0$

Coordinate Geometry - Circle

Prove that the length of the common chord of the circles $\left(x - a\right)^2 + \left(y - b\right)^2 = c^2$ and $\left(x - b\right)^2 + \left(y - a\right)^2 = c^2$ is $\sqrt{4c^2 - 2 \left(a - b\right)^2}$.

Given circles:

$\left(x - a\right)^2 + \left(y - b\right)^2 = c^2$

i.e. $\;$ $x^2 + y^2 -2ax -2by + a^2 + b^2 - c^2 = 0$ $\;\;\; \cdots \; (1)$

$\left(x - b\right)^2 + \left(y - a\right)^2 = c^2$

i.e. $\;$ $x^2 + y^2 -2bx - 2ay + a^2 + b^2 - c^2 = 0$ $\;\;\; \cdots \; (2)$

Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$ $\;$ gives

$g_1 = -a, \; f_1 = -b, \; c_1 = a^2 + b^2 - c^2$

Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ $\;$ gives

$g_2 = -b, \; f_2 = -a, \; c_2 = a^2 + b^2 - c^2$

Common chord of circles $(1)$ and $(2)$ is

$2x \left(g_1 - g_2\right) + 2y \left(f_1 - f_2\right) + c_1 - c_2 = 0$

i.e. $\;$ $2x \left(-a + b\right) + 2y \left(-b + a\right) + a^2 + b^2 - c^2 - a^2 - b^2 + c^2 = 0$

i.e. $\;$ $2 \left(b - a\right)x + 2 \left(a - b\right) y = 0$

i.e. $\;$ $\left(b - a\right) x = \left(b - a\right) y$

i.e. $\;$ $x = y$ $\;\;\; \cdots \; (3)$ $\;\;$ provided $b - a \neq 0$

Substituting $x = y$ in equation $(1)$ gives

$y^2 + y^2 - 2ay - 2by + a^2 + b^2 - c^2 = 0$

i.e. $\;$ $2y^2 - \left(2a + 2b\right) y + a^2 + b^2 - c^2 = 0$

$\begin{aligned} \therefore \; y & = \dfrac{\left(2a + 2b\right) \pm \sqrt{\left(2a + 2b\right)^2 - 4 \times 2 \times \left(a^2 + b^2 - c^2\right)}}{2 \times 2} \\\\ & = \dfrac{\left(2a + 2b\right) \pm \sqrt{4a^2 + 4b^2 + 8ab - 8a^2 - 8b^2 + 8c^2}}{4} \\\\ & = \dfrac{\left(2a + 2b\right) \pm \sqrt{8c^2 - \left(4a^2 + 4b^2 - 8ab\right)}}{4} \\\\ & = \dfrac{2 \left(a + b\right) \pm 2 \sqrt{2c^2 - \left(a^2 + b^2 - 2ab\right)}}{4} \\\\ & = \dfrac{a + b \pm \sqrt{2c^2 - \left(a - b\right)^2}}{2} \end{aligned}$

$\therefore \;$ From equation $(3)$,

when $\;$ $y = \dfrac{a + b + \sqrt{2c^2 - \left(a - b\right)^2}}{2}$, $\;$ $x = \dfrac{a + b + \sqrt{2c^2 - \left(a - b\right)^2}}{2}$ $\;$ and

when $\;$ $y = \dfrac{a + b - \sqrt{2c^2 - \left(a - b\right)^2}}{2}$, $\;$ $x = \dfrac{a + b - \sqrt{2c^2 - \left(a - b\right)^2}}{2}$

$\therefore \;$ Coordinates of the end points of the common chord of the two circles are

$P \left(x_1, y_1\right) = \left(\dfrac{a + b + \sqrt{2c^2 - \left(a - b\right)^2}}{2}, \dfrac{a + b + \sqrt{2c^2 - \left(a - b\right)^2}}{2}\right)$ $\;$ and

$Q \left(x_2, y_2\right) = \left(\dfrac{a + b - \sqrt{2c^2 - \left(a - b\right)^2}}{2}, \dfrac{a + b - \sqrt{2c^2 - \left(a - b\right)^2}}{2}\right)$

$\therefore \;$ Square of length of common chord of the two given circles is

$PQ^2 = \left(\dfrac{a + b}{2} + \dfrac{\sqrt{2c^2 - \left(a - b\right)^2}}{2} - \dfrac{a + b}{2} + \dfrac{\sqrt{2c^2 - \left(a - b\right)^2}}{2}\right)^2$

$\hspace{1.5cm}$ $+ \left(\dfrac{a + b}{2} + \dfrac{\sqrt{2c^2 - \left(a - b\right)^2}}{2} - \dfrac{a + b}{2} + \dfrac{\sqrt{2c^2 - \left(a - b\right)^2}}{2}\right)^2$

i.e. $\;$ $PQ^2 = 2c^2 - \left(a - b\right)^2 + 2c^2 - \left(a - b\right)^2$

i.e. $\;$ $PQ^2 = 4c^2 - 2 \left(a - b\right)^2$

$\therefore \;$ Length of common chord of the two given circles is

$PQ = \sqrt{4c^2 - 2 \left(a - b\right)^2}$

Hence proved.

Coordinate Geometry - Circle

Find the equation of the circle whose diameter is the common chord of the circles $x^2 + y^2 + 2x + 3y + 1 = 0$ and $x^2 + y^2 + 4x + 3y + 2 = 0$

Given circles:

$x^2 + y^2 + 2x + 3y + 1 = 0$ $\;\;\; \cdots \; (1)$

$x^2 + y^2 + 4x + 3y + 2 = 0$ $\;\;\; \cdots \; (2)$

Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$ $\;$ gives

$g_1 = 1, \; f_1 = \dfrac{3}{2}, \; c_1 = 1$

Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ $\;$ gives

$g_2 = 2, \; f_2 = \dfrac{3}{2}, \; c_2 = 2$

Common chord of circles $(1)$ and $(2)$ is

$2x \left(g_1 - g_2\right) + 2y \left(f_1 - f_2\right) + c_1 - c_2 = 0$

i.e. $\;$ $2x \left(1 - 2\right) + 2y \left(\dfrac{3}{2} - \dfrac{3}{2}\right) + 1 - 2 = 0$

i.e. $\;$ $-2x -1 = 0$

i.e. $\;$ $x = \dfrac{-1}{2}$

Substituting the value of $x$ in equation $(1)$ gives

$\dfrac{1}{4} + y^2 - 1 + 3y + 1 = 0$

i.e. $\;$ $y^2 + 3y + \dfrac{1}{4} = 0$

i.e. $\;$ $4y^2 + 12 y + 1 = 0$

Solving the quadratic equation gives $\;$ $y = \dfrac{-3}{2} \pm \sqrt{2}$

$\therefore \;$ As per question, the coordinates of the end points of the diameter of the required circle are

$A \left(x_1, y_1\right) = \left(\dfrac{-1}{2}, \dfrac{-3}{2} + \sqrt{2}\right)$ $\;$ and $\;$ $B \left(x_2, y_2\right) = \left(\dfrac{-1}{2}, \dfrac{-3}{2} - \sqrt{2}\right)$

$\therefore \;$ Equation equation of the required circle is

$\left(x - x_1\right) \left(x - x_2\right) + \left(y - y_1\right) \left(y - y_2\right) = 0$

i.e. $\;$ $\left(x - \dfrac{1}{2}\right) \left(x - \dfrac{1}{2}\right) + \left(y + \dfrac{3}{2} - \sqrt{2}\right) \left(y + \dfrac{3}{2} + \sqrt{2}\right) = 0$

i.e. $\;$ $x^2 - x + \dfrac{1}{4} + y^2 + \dfrac{9}{4} + 3y - 2 = 0$

i.e. $\;$ $x^2 + y^2 - x + 3y + \dfrac{1}{2} = 0$

i.e. $\;$ $2x^2 + 2y^2 - 2x + 6y + 1 = 0$

Coordinate Geometry - Circle

Find the equation of a system of circles which have the line $x - y = 0$ for their radical axis.

Equation of system of circles is

$x^2 + y^2 + 2gx + 2fy + c + \lambda \left(px + qy + r\right) = 0$ $\;\;\; \cdots \; (1)$

where $\;$ $\lambda$ $\;$ is a constant and

$px + qy + r = 0$ $\;\;\; \cdots \; (2)$ $\;$ is the radical axis.

Given: $\;$ Radical axis: $\;$ $x - y = 0$ $\;\;\; \cdots \; (3)$

Since equations $(2)$ and $(3)$ represent the same line,

coefficients of the $x$ and $y$ terms and the constant term must be identical.

i.e. $\;$ $p = 1, \; q = -1, \; r = 0$

$\therefore \;$ Substituting the values of $p$, $q$ and $r$ in equation $(1)$, the required equation of system of circles is

$x^2 + y^2 + 2gx + 2fy + c + \lambda \left(x - y\right) = 0$

i.e. $\;$ $x^2 + y^2 + \left(2g + \lambda\right) x + \left(2f - \lambda\right) y + c = 0$

Coordinate Geometry - Circle

Find the equation of the circle which has extremities of a diameter the origin and the point $\left(2, -4\right)$.
Find also the equations of the tangents to the circle which are parallel to this diameter.

The extremities of the diameter are

origin, $\;$ i.e. $\;$ $\left(x_1, y_1\right) = \left(0, 0\right)$ $\;$ and $\;$ $\left(x_2, y_2\right) = \left(2, -4\right)$

$\therefore \;$ Equation of required circle is

$\left(x - x_1\right) \left(x - x_2\right) + \left(y - y_1\right) \left(y - y_2\right) = 0$

i.e. $\;$ $\left(x - 0\right) \left(x - 2\right) + \left(y - 0\right) \left(y + 4\right) = 0$

i.e. $\;$ $x^2 + y^2 - 2x + 4y = 0$ $\;\;\; \cdots \; (1)$

Slope of the line passing through the points $\;$ $\left(0, 0\right)$ $\;$ and $\;$ $\left(2, -4\right)$ $\;$ is

$m = \dfrac{-4}{2} = -2$

Since the required tangents are parallel to the line through $\left(0, 0\right)$ and $\left(2, -4\right)$,

$\therefore \;$ Slope of required tangents $= m = -2$

Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$g = -1, \; f = 2, \; c = 0$

$\therefore \;$ Radius of circle $(1)$ is

$a = \sqrt{g^2 + f^2 - c} = \sqrt{\left(-1\right)^2 + 2^2 - 0} = \sqrt{5}$

Equations of required tangents are

$y = mx \pm a \sqrt{1 + m^2}$

i.e. $\;$ $y = -2x \pm \sqrt{5} \times \sqrt{1 + \left(-2\right)^2}$

i.e. $\;$ $y = -2x \pm 5$

$\therefore \;$ The equations of the required tangents are

$2x + y + 5 = 0$ $\;$ and $\;$ $2x + y - 5 = 0$

Coordinate Geometry - Circle

Find the equations of the tangents to the circle $\;$ $2x^2 + 2y^2 = 5$ $\;$ which are perpendicular to $\;$ $y = 2x$.

Given circle: $\;\;$ $2x^2 + 2y^2 = 5$

i.e. $\;$ $x^2 + y^2 = \dfrac{5}{2}$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 = a^2$ $\;$ gives

$a^2 = \dfrac{5}{2}$ $\implies$ $a = \pm \sqrt{\dfrac{5}{2}}$

Given line: $\;\;$ $y = 2x$ $\;\;\; \cdots \; (2)$

Slope of the given line $= m_1 = 2$

Since the required tangents are perpendicular to $(2)$,

$\therefore \;$ slope of tangents $= m = \dfrac{-1}{m_1} = \dfrac{-1}{2}$

Equations of tangents are

$y = mx \pm a \sqrt{1 + m^2}$

i.e. $\;$ $y = \dfrac{-1}{2}x \pm \sqrt{\dfrac{5}{2}} \times \sqrt{1 + \dfrac{1}{4}}$

i.e. $\;$ $y = \dfrac{-1}{2}x \pm \sqrt{\dfrac{5}{2}} \times \dfrac{\sqrt{5}}{2}$

i.e. $\;$ $y = \dfrac{-1}{2} x \pm \dfrac{5}{2 \sqrt{2}}$

i.e. $\;$ $2 \sqrt{2} y = - \sqrt{2} x \pm 5$

$\therefore \;$ The equations of the required tangents are

$\sqrt{2} x + 2 \sqrt{2} y + 5 = 0$ $\;$ and $\;$ $\sqrt{2} x + 2 \sqrt{2} y - 5 = 0$

Coordinate Geometry - Circle

Find the circle which cuts the circles $\;$ $x^2 + \left(y - b\right)^2 = a^2$, $\;$ $\left(x - c\right)^2 + y^2 = a^2$ $\;$ and $\;$ $x^2 + y^2 = a^2$ $\;$ orthogonally.

Given circles:

$x^2 + \left(y - b\right)^2 = a^2$

i.e. $\;$ $x^2 + y^2 - 2by + b^2 - a^2 = 0$ $\;\;\; \cdots \; (1)$

$\left(x - c\right)^2 + y^2 = a^2$

i.e. $\;$ $x^2 + y^2 -2cx + c^2 - a^2 = 0$ $\;\;\; \cdots \; (2)$

$x^2 + y^2 = a^2$ $\;\;\; \cdots \; (3)$

Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ $\;$ gives

$g_1 = 0, \; f_1 = -b, \; c_1 = b^2 - a^2$

Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ $\;$ gives

$g_2 = -c, \; f_2 = 0, \; c_2 = c^2 - a^2$

Comparing equation $(3)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_3 x + 2f_3 y + c_3 = 0$ $\;$ gives

$g_3 = 0, \; f_3 = 0, \; c_3 = - a^2$

The radical axis of circles $(1)$ and $(2)$ is

$2 x \left(g_1 - g_2\right) + 2y \left(f_1 - f_2\right) + c_1 - c_2 = 0$

i.e. $\;$ $2x \left(0 + c\right) + 2y \left(-b - 0\right) + b^2 - a^2 - c^2 + a^2 = 0$

i.e. $\;$ $2cx - 2by + b^2 - c^2 = 0$ $\;\;\; \cdots \; (4)$

The radical axis of circles $(1)$ and $(3)$ is

$2 x \left(g_1 - g_3\right) + 2y \left(f_1 - f_3\right) + c_1 - c_3 = 0$

i.e. $\;$ $2x \times 0 + 2y \left(-b - 0\right) + b^2 - a^2 + a^2 = 0$

i.e. $\;$ $-2by + b^2 = 0$ $\;\;\; \cdots \; (5)$

From equation $(5)$, if $\;$ $b \neq 0$, $\;$ then $\;$ $y = \dfrac{b}{2}$

Substituting $\;$ $y = \dfrac{b}{2}$ $\;$ in equation $(4)$ gives,

$2cx - 2 b \times \dfrac{b}{2} + b^2 - c^2 = 0$

i.e. $\;$ $2 cx - c^2 = 0$

$\implies$ $x = \dfrac{c}{2}$ $\;$ if $\;$ $c \neq 0$

$\therefore \;$ The radical center is $\;$ $\left(h, k\right) = \left(\dfrac{c}{2}, \dfrac{b}{2}\right)$

Square of tangent from $\left(\dfrac{c}{2}, \dfrac{b}{2}\right)$ to equation $(1)$ is

$h^2 + k^2 + 2 g_1 h + 2 f_1 k + c_1$

$= \dfrac{c^2}{4} + \dfrac{b^2}{4} + 2 \times 0 \times \dfrac{c}{2} + 2 \times \left(-b\right) \times \dfrac{b}{2} + b^2 - a^2$

$= \dfrac{c^2}{4} + \dfrac{b^2}{4} - a^2$

$= \dfrac{b^2 + c^2 - 4 a^2}{4}$

$\therefore \;$ Required equation of circle is

$\left(x - \dfrac{c}{2}\right)^2 + \left(y - \dfrac{b}{2}\right)^2 = \dfrac{b^2 + c^2 - 4a^2}{4}$

i.e. $\;$ $x^2 - cx + \dfrac{c^2}{4} + y^2 - by + \dfrac{b^2}{4} = \dfrac{b^2}{4} + \dfrac{c^2}{4} - a^2$

i.e. $\;$ $x^2 + y^2 - cx - by + a^2 = 0$

Coordinate Geometry - Circle

Find the radical center of the circles $\;$ $x^2 + y^2 + 2x + 2y - 2 = 0$, $\;$ $x^2 + y^2 -4x - 6y + 6 = 0$ $\;$ and $\;$ $x^2 + y^2 + 6x - 4y - 12 = 0$.

Given circles: $\;$ $x^2 + y^2 + 2x + 2y - 2 = 0$ $\;\;\; \cdots \; (1)$

$x^2 + y^2 - 4x - 6y + 6 = 0$ $\;\;\; \cdots \; (2)$

$x^2 + y^2 + 6x - 4y - 12 = 0$ $\;\;\; \cdots \; (3)$

Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$g = 1, \; f = 1, \; c = -2$

Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$ $\;$ gives

$g_1 = -2, \; f_1 = -3, \; c_1 = 6$

Comparing equation $(3)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ $\;$ gives

$g_2 = 3, \; f_2 = -2, \; c_2 = - 12$

The radical axis of circles $(1)$ and $(2)$ is

$2 x \left(g - g_1\right) + 2y \left(f - f_1\right) + c - c_1 = 0$

i.e. $\;$ $2x \left(1 + 2\right) + 2y \left(1 + 3\right) - 2 - 6 = 0$

i.e. $\;$ $6x + 8 y - 8 = 0$

i.e. $\;$ $3x + 4y - 4 = 0$ $\;\;\; \cdots \; (4)$

The radical axis of circles $(1)$ and $(3)$ is

$2 x \left(g - g_2\right) + 2y \left(f - f_2\right) + c - c_2 = 0$

i.e. $\;$ $2x \left(1 - 3\right) + 2y \left(1 + 2\right) - 2 + 12 = 0$

i.e. $\;$ $- 4x + 6y + 10 = 0$

i.e. $\;$ $-2x + 3y + 5 = 0$ $\;\;\; \cdots \; (5)$

Solving equations $(4)$ and $(5)$ simultaneously gives,

$x = \dfrac{32}{17}, \; y = \dfrac{-7}{17}$

$\therefore \;$ The radical center is $\;$ $\left(\dfrac{32}{17}, \dfrac{-7}{17}\right)$

Coordinate Geometry - Circle

Find the radical axis of the circles $x^2 + y^2 - 4x + 6y - 10 = 0$ and $x^2 + y^2 + 2x - 6y + 2 = 0$ and show that the circles intersect in real points.

Given circles: $\;$ $x^2 + y^2 - 4x + 6y - 10 = 0$ $\;\;\; \cdots \; (1)$

$x^2 + y^2 + 2x - 6y + 2 = 0$ $\;\;\; \cdots \; (2)$

Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$g = -2, \; f = 3, \; c = -10$

Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$ $\;$ gives

$g_1 = 1, \; f_1 = -3, \; c_1 = 2$

The radical axis of circles $(1)$ and $(2)$ is

$2 x \left(g - g_1\right) + 2y \left(f - f_1\right) + c - c_1 = 0$

i.e. $\;$ $2x \left(-2-1\right) + 2y \left(3 + 3\right) - 10 - 2 = 0$

i.e. $\;$ $-6x + 12 y - 12 = 0$

i.e. $\;$ $x - 2y + 2 = 0$ $\;\;\; \cdots \; (3)$

Let circles $(1)$ and $(2)$ have a common point $\left(x_1, y_1\right)$

Then, $\;$ $x_1^2 + y_1^2 - 4 x_1 + 6 y_1 - 10 = 0$

and $\;$ $x_1^2 + y_1^2 + 2x_1 - 6 y_1 + 2 = 0$

$\therefore \;$ By subtraction, $\;$ $- 6x_1 + 12 y_1 - 12 = 0$

i.e. $\;$ $x_1 - 2 y_1 + 2 = 0$

$\therefore \;$ The equation of the common chord of the two circles is

$x - 2y + 2 = 0$ $\;\;\; \cdots \; (4)$

which is the same as equation $(3)$.

i.e. $\;$ Radical axis is the common chord of the circles.

$\implies$ The two circles intersect in real points.

Coordinate Geometry - Circle

The polar of the point $P \left(x_1, y_1\right)$ with respect to the circle $x^2 + y^2 = a^2$ meets the axes at $Q$ and $R$. Prove that the area of $\triangle OQR$ is $\dfrac{a^4}{2 x_1 y_1}$ where $O$ is the origin.

Given point: $\;$ $P \left(x_1, y_1\right)$

Given circle: $\;$ $x^2 + y^2 = a^2$

The polar of $P \left(x_1, y_1\right)$ with respect to the circle $\;$ $x^2 + y^2 = a^2$ $\;$ is

$x x_1 + y y_1 = a^2$ $\;\;\; \cdots \; (1)$

Let equation $(1)$ meet the $X$ axis at $Q$ and the $Y$ axis at $R$.

Then, $\;$ $Q = \left(\dfrac{a^2}{x_1}, 0\right)$ $\;$ and $\;$ $R = \left(0, \dfrac{a^2}{y_1}\right)$

Area of $\triangle OQR$

$= \dfrac{1}{2} \times OQ \times OR$

$= \dfrac{1}{2} \times \dfrac{a^2}{x_1} \times \dfrac{a^2}{y_1}$

$= \dfrac{a^4}{2 x_1 y_1}$

Coordinate Geometry - Circle

If the pole of a straight line with respect to the circle $x^2 + y^2 = r^2$ lies on the circle $x^2 + y^2 = k^2 r^2$, prove that the line will touch the circle $x^2 + y^2 = \dfrac{r^2}{k^2}$.

Let the straight line be $\;\;$ $ax + by + c = 0$ $\;\;\; \cdots \; (1)$

Equation of circle $\;\;$ $x^2 + y^2 = r^2$ $\;\;\; \cdots \; (2)$

Let $\left(h, k\right)$ be the pole of equation $(1)$ with respect to equation $(2)$.

Then, equation $(1)$ must be identical with the polar of $\left(h, k\right)$ with respect to $(2)$.

i.e. $\;$ with $\;$ $xh + yk = r^2$

i.e. $\;$ with $\;$ $xh + yk - r^2 = 0$ $\;\;\; \cdots \; (3)$

Given: $\;$ $\left(h, k\right)$ lies on the circle $\;$ $x^2 + y^2 = k^2 r^2$

$\therefore \;$ We have, $\;$ $h^2 + k^2 = k^2 r^2$

$\implies$ $h^2 = k^2 \left(r^2 - 1\right)$ $\;\;\; \cdots \; (4)$

Now, equation $(3)$ can be written as

$y = \dfrac{-h}{k} x + \dfrac{r^2}{k}$ $\;\;\; \cdots \; (5)$

Comparing equation $(5)$ with the standard equation of line $\;$ $y = mx + c_1$ $\;$ gives

slope $= m = \dfrac{-h}{k}$ $\;\;\; \cdots \; (6a)$

and $\;$ intercept $= c_1 = \dfrac{r^2}{k}$ $\;\;\; \cdots \; (6b)$

Given equation of circle $\;\;$ $x^2 + y^2 = \dfrac{r^2}{k^2}$ $\;\;\; \cdots \; (7)$

Comparing equation $(7)$ with the standard equation of circle $\;$ $x^2 + y^2 = R^2$ $\;$ gives $\;\;$ $R^2 = \dfrac{r^2}{k^2}$

The line $\;$ $y = mx + c_1$ $\;$ will touch the circle $\;$ $x^2 + y^2 = R^2$ $\;$ if $\;\;$ $c_1^2 = R^2 \left(1 + m^2\right)$

$\therefore \;$ Equation $(5)$ will touch equation $(7)$

if $\;$ $\left(\dfrac{r^2}{k}\right)^2 = \dfrac{r^2}{k^2} \left(1 + \dfrac{h^2}{k^2}\right)$

i.e. $\;$ if $\;$ $\dfrac{r^4}{k^2} = \dfrac{r^2}{k^2} + \dfrac{r^2 h^2}{k^4}$ $\;\;\; \cdots \; (8)$

Now, in view of equation $(4)$, equation $(8)$ becomes,

if $\;$ $\dfrac{r^4}{k^2} = \dfrac{r^2}{k^2} + \dfrac{r^2}{k^4} \times k^2 \left(r^2 - 1\right)$

i.e. $\;$ if $\;$ $\dfrac{r^4}{k^2} = \dfrac{r^2}{k^2} + \dfrac{r^2}{k^2} \left(r^2 - 1\right)$

i.e. $\;$ if $\;$ $\dfrac{r^4}{k^2} = \dfrac{r^2}{k^2} + \dfrac{r^4}{k^2} - \dfrac{r^2}{k^2}$

i.e. $\;$ if $\;$ $\dfrac{r^4}{k^2} = \dfrac{r^4}{k^2}$ $\;\;$ which is true.

$\implies$ Equation $(5)$ touches circle $(7)$.

But, equations $(5)$ and $(1)$ are identical.

$\implies$ Equation $(1)$ touches circle $(7)$.

Hence proved.

Coordinate Geometry - Circle

Find the pole of the straight line $\;$ $x - y + 2 = 0$ $\;$ with respect to the circle $\;$ $x^2 + y^2 - 4x + 6y - 12 = 0$.

Given line: $\;$ $x - y + 2 = 0$ $\;\;\; \cdots \; (1)$

Given circle: $\;$ $x^2 + y^2 - 4x + 6y - 12 = 0$ $\;\;\; \cdots \; (2)$

Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$g = -2, \; f = 3, \; c = -12$

Let $\;$ $\left(h, k\right)$ $\;$ be the pole of equation $(1)$ with respect to equation $(2)$.

Then equation $(1)$ must be identical with the polar of $\;$ $\left(h, k\right)$ $\;$ with respect to equation $(2)$

i.e. $\;$ with $\;$ $xh + yk + g \left(x + h\right) + f \left(y + k\right) + c = 0$

i.e. $\;$ with $\;$ $xh + yk - 2 \left(x + h\right) + 3 \left(y + k\right) - 12 = 0$

i.e. $\;$ with $\;$ $x \left(h - 2\right) + y \left(k + 3\right) - 2h + 3k - 12 = 0$ $\;\;\; \cdots \; (3)$

$\therefore \;$ Comparing equations $(1)$ and $(3)$ gives

$\dfrac{h - 2}{1} = \dfrac{k + 3}{-1} = \dfrac{-2h + 3k - 12}{2}$

Solving, we get,

$2h - 4 = -2h + 3k - 12$

i.e. $\;$ $4h - 3k + 8 = 0$ $\;\;\; \cdots \; (4)$

and $\;$ $2k + 6 = 2h - 3k + 12$

i.e. $\;$ $2h - 5k + 6 = 0$ $\;\;\; \cdots \; (5)$

Solving equations $(4)$ and $(5)$ simultaneously gives

$h = \dfrac{-11}{7}, \; k = \dfrac{4}{7}$

$\therefore \;$ The pole of equation $(1)$ with respect to equation $(2)$ is $\left(\dfrac{-11}{7}, \dfrac{4}{7}\right)$.

Coordinate Geometry - Circle

Find the pole of the straight line $\;$ $3x + 4y - 12 = 0$ $\;$ with respect to the circle $\;$ $x^2 + y^2 = 24$.

Given line: $\;$ $3x + 4y - 12 = 0$ $\;\;\; \cdots \; (1)$

Given circle: $\;$ $x^2 + y^2 = 24$ $\;\;\; \cdots \; (2)$

Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 = r^2$

gives $\;$ $r^2 = 24$

Let $\;$ $\left(h, k\right)$ $\;$ be the pole of equation $(1)$ with respect to equation $(2)$.

Then equation $(1)$ must be identical with the polar of $\;$ $\left(h, k\right)$ $\;$ with respect to equation $(2)$

i.e. $\;$ with $\;$ $xh + yk = r^2$

i.e. $\;$ with $\;$ $xh + yk = 24$

i.e. $\;$ with $\;$ $xh + yk - 24 = 0$ $\;\;\; \cdots \; (3)$

$\therefore \;$ Comparing equations $(1)$ and $(3)$ gives

$\dfrac{h}{3} = \dfrac{k}{4} = \dfrac{-24}{-12}$

Solving, we get,

$h = 6, \; k = 8$

$\therefore \;$ The pole of equation $(1)$ with respect to equation $(2)$ is $\left(6, 8\right)$.

Coordinate Geometry - Circle

Find the polar of the point $\left(h, k\right)$ with respect to the circle $x^2 + y^2 - 2gx - 2fy + c = 0$.

Given point: $\;\;$ $\left(h, k\right)$

Given circle: $\;\;$ $x^2 + y^2 - 2gx - 2fy + c = 0$

Comparing the equation of the given circle with the standard equation of circle $\;\;$ $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$

gives $\;\;$ $g_1 = -g, \; f_1 = -f, \; c_1 = c$

The polar of $\;$ $\left(h, k\right)$ $\;$ with respect to the circle $\;$ $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ $\;$ is

$xh + yk + g_1 \left(x + h\right) + f_1 \left(y + k\right) + c_1 = 0$

$\therefore \;$ The required polar is:

$xh + yk -g \left(x + h\right) -f \left(y + k\right) + c = 0$

i.e. $\;$ $xh + yk -gx -gh -fy -fk + c = 0$

i.e. $\;$ $x \left(h - g\right) + y \left(k - f\right) -gh -fk + c = 0$

Coordinate Geometry - Circle

Find the polar of the point $\left(-7, -9\right)$ with respect to the circle $x^2 + y^2 - 12x - 8y - 48 = 0$.

Given point: $\;\;$ $\left(h, k\right) = \left(-7, -9\right)$

Given circle: $\;\;$ $x^2 + y^2 - 12x - 8y - 48 = 0$

Comparing the equation of the given circle with the standard equation of circle $\;\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$

gives $\;\;$ $g = -6, \; f = -4, \; c = -48$

The polar of $\;$ $\left(h, k\right)$ $\;$ with respect to the circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ is $\;$ $xh + yk + g \left(x + h\right) + f \left(y + k\right) + c = 0$

$\therefore \;$ The required polar is:

$-7x -9y - 6 \left(x - 7\right) - 4 \left(y - 9\right) - 48 = 0$

i.e. $\;$ $-7x -9y - 6x + 42 - 4y + 36 - 48 = 0$

i.e. $\;$ $-13x - 13y + 30 = 0$

Coordinate Geometry - Circle

Find the polar of the point $\left(3, 4\right)$ with respect to the circle $x^2 + y^2 = 25$.

Given point: $\;\;$ $\left(h, k\right) = \left(3, 4\right)$

Given circle: $\;\;$ $x^2 + y^2 = 25$

Comparing the equation of the given circle with the standard equation of circle $\;\;$ $x^2 + y^2 = r^2$

gives $\;\;$ $r^2 = 25$

The polar of $\;$ $\left(h, k\right)$ $\;$ with respect to the circle $\;$ $x^2 + y^2 = r^2$ $\;$ is $\;$ $xh + yk = r^2$

$\therefore \;$ The required polar is: $\;$ $3x + 4y = 25$

Coordinate Geometry - Circle

Find the angle between the tangents drawn from $\left(3, 4\right)$ to the circle $x^2 + y^2 - 4x - 2y - 4 = 0$.

External point: $\;$ $P \left(3, 4\right)$

Equation of circle: $\;$ $x^2 + y^2 - 4x - 2y - 4 = 0$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of a circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$g = -2, \; f = -1, \; c = -4$

$\therefore \;$ Center of circle $= C = \left(-g, -f\right) = \left(2, 1\right)$

Radius of circle $= r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 1 + 4} = 3$

Let $\theta$ be the angle between the two tangents drawn from the external point $P$.

Then, $\;\;$ $\sin \left(\dfrac{\theta}{2}\right) = \dfrac{r}{CP}$

i.e. $\;$ $\sin \left(\dfrac{\theta}{2}\right) = \dfrac{3}{\sqrt{\left(3 - 2\right)^2 + \left(4 - 1\right)^2}} = \dfrac{3}{\sqrt{1 + 9}} = \dfrac{3}{\sqrt{10}}$

i.e. $\;$ $\dfrac{\theta}{2} = \sin^{-1} \left(\dfrac{3}{\sqrt{10}}\right)$

i.e. $\;$ $\theta = 2 \sin^{-1} \left(\dfrac{3}{\sqrt{10}}\right)$

Coordinate Geometry - Circle

Find the condition that the straight line $\;$ $x \cos \alpha + y \sin \alpha - p = 0$ $\;$ should touch the circle $\;$ $x^2 + y^2 - 2 \mu x = a^2$, $\;$ $\mu$ being a variable quantity. If $\mu_1$ and $\mu_2$ be the values of $\mu$, show that $\;$ $\mu_1 \mu_2 = \left(a^2 - p^2\right) \text{cosec}^2 \alpha$

Given circle: $\;$ $x^2 + y^2 - 2 \mu x = a^2$

i.e. $\;$ $x^2 + y^2 - 2 \mu x - a^2 = 0$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$g = - \mu, \; f = 0, \; c = - a^2$

$\therefore \;$ Center of the circle $= \left(-g, -f\right) = \left(\mu, 0\right)$

Radius of the circle $= r = \sqrt{g^2 + f^2 - c} = \sqrt{\mu^2 + a^2}$

Given line: $\;$ $x \cos \alpha + y \sin \alpha - p = 0$ $\;\;\; \cdots \; (2)$

Given line [equation $(2)$] is a tangent to the given circle [equation $(1)$] if

distance of the line from the center $=$ radius of the circle

i.e. $\;$ $\left|\dfrac{\mu \cos \alpha + 0 - p}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}}\right| = \sqrt{\mu^2 + a^2}$

i.e. $\;$ $\mu^2 \cos^2 \alpha + p^2 - 2 p \mu \cos \alpha = \mu^2 + a^2$

i.e. $\;$ $\mu^2 \left(\cos^2 \alpha - 1\right) - 2 p \mu \cos \alpha + p^2 - a^2 = 0$

i.e. $\;$ $- \mu^2 \sin^2 \alpha - 2 p \mu \cos \alpha + p^2 - a^2 = 0$

i.e. $\;$ $\mu^2 + 2 p \mu \dfrac{\cos \alpha}{\sin^2 \alpha} + \dfrac{a^2 - p^2}{\sin^2 \alpha} = 0$

i.e. $\;$ $\mu^2 + 2 p \mu \cot \alpha \text{ cosec} \alpha + \left(a^2 - p^2\right) \text{cosec}^2 \alpha = 0$ $\;\;\; \cdots \; (3)$

Equation $(3)$ is the required condition to be satisfied so that line $(2)$ is a tangent to the circle $(1)$.

Equation $(3)$ is a quadratic equation in $\mu$.

If $\mu_1, \; \mu_2$ are the values of $\mu$, then

Product of roots of equation $(3)$ $= \mu_1 \mu_2 = \dfrac{\left(a^2 - p^2\right) \text{cosec}^2 \alpha}{1} = \left(a^2 - p^2\right) \text{cosec}^2 \alpha$

Coordinate Geometry - Circle

Find the equation of the circle which cuts the circles $\;$ $x^2 + y^2 - 2x + 3y + 4 = 0$ $\;$ and $\;$ $x^2 + y^2 + 3x - 5y + 1 = 0$ $\;$ orthogonally and which passes through the origin.

Given circles: $\;$ $x^2 + y^2 - 2x + 3y + 4 = 0$ $\;\;\; \cdots \; (1a)$

and $\;$ $x^2 + y^2 + 3x - 5y + 1 = 0$ $\;\;\; \cdots \; (1b)$

Comparing equation $(1a)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$ $\;\;\; \cdots \; (2a)$ $\;$ gives

$g_1 = -1, \; f_1 = \dfrac{3}{2}, \; c_1 = 4$

Comparing equation $(1b)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ $\;\;\; \cdots \; (2b)$ $\;$ gives

$g_2 = \dfrac{3}{2}, \; f_2 = \dfrac{-5}{2}, \; c_2 = 1$

Let the equation of the required circle be

$x^2 + y^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (2a)$

$\because \;$ Equation $(2a)$ passes through the origin $\left(0, 0\right)$, we have $\;\;\;$ $c = 0$

$\therefore \;$ The equation of the required circle becomes

$x^2 + y^2 + 2gx + 2fy = 0$ $\;\;\; \cdots \; (2b)$

$\because \;$ Circles $(1a)$ and $(2b)$ are orthogonal, we have,

$2gg_1 + 2ff_1 = c + c_1$

i.e. $\;$ $2 \times g \times \left(-1\right) + 2 \times f \times \dfrac{3}{2} = 0 + 4$

i.e. $\;$ $-2g + 3f = 4$ $\;\;\; \cdots \; (3a)$

$\because \;$ Circles $(1b)$ and $(2b)$ are orthogonal, we have,

$2 g g_2 + 2 f f_2 = c + c_2$

i.e. $\;$ $2 \times g \times \dfrac{3}{2} + 2 \times f \times \left(\dfrac{-5}{2}\right) = 0 + 1$

i.e. $\;$ $3g - 5f = 1$ $\;\;\; \cdots \; (3b)$

Solving equations $(3a)$ and $(3b)$ simultaneously gives

$g = -23, \; f = -14$

$\therefore \;$ Equation of the required circle is

$x^2 + y^2 - 46x - 28y = 0$

Coordinate Geometry - Circle

Prove that the circles $\;$ $x^2 + y^2 + 2ax +c = 0$ $\;$ and $\;$ $x^2 + y^2 - 2by - c = 0$ $\;$ cut orthogonally.

Given circles: $\;$ $x^2 + y^2 + 2ax + c = 0$ $\;\;\; \cdots \; (1a)$

and $\;$ $x^2 + y^2 - 2by - c = 0$ $\;\;\; \cdots \; (1b)$

Comparing equation $(1a)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$ $\;\;\; \cdots \; (2a)$ $\;$ gives

$g_1 = a, \; f_1 = 0, \; c_1 = c$

Comparing equation $(1b)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ $\;\;\; \cdots \; (2b)$ $\;$ gives

$g_2 = 0, \; f_2 = -b, \; c_2 = -c$

Circles $(2a)$ and $(2b)$ are orthogonal if

$2g_1 g_2 + 2f_1 f_2 = c_1 + c_2$

$\therefore \;$ We have

$2 \times a \times 0 + 2 \times 0 \times \left(-b\right) = c - c$

i.e. $\;$ $0 = 0$ $\;\;$ which is true.

$\implies$ Circles $(1a)$ and $(1b)$ are orthogonal.

Coordinate Geometry - Circle

Find the equation of the circle on the line joining the points $\left(1, 5\right)$, $\;$ $\left(3, 1\right)$ as diameter and find the extremities of the perpendicular diameter.

Given points: $\;$ $A\left(x_1, y_1\right) = \left(1, 5\right)$, $\;$ $B \left(x_2, y_2\right) = \left(3, 1\right)$

Equation of circle in terms of ends of diameter $A \left(x_1, y_1\right)$ and $B \left(x_2, y_2\right)$ is

$\left(x - x_1\right) \left(x - x_2\right) + \left(y - y_1\right) \left(y - y_2\right) = 0$

$\therefore \;$ The required equation of circle is

$\left(x - 1\right) \left(x - 3\right) + \left(y - 5\right) \left(y - 1\right) = 0$

i.e. $\;$ $x^2 - 4x + 3 + y^2 - 6y + 5 = 0$

i.e. $\;$ $x^2 + y^2 - 4x - 6y + 8 = 0$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$g = -2, \; f = -3, \; c = 8$

$\therefore \;$ Center of circle given by equation $(1)$ is

$C \left(-g, -f\right) = \left(2, 3\right)$

Slope of diameter $AB = \dfrac{1 - 5}{3 - 1} = -2$

Let $CD$ be the diameter perpendicular to $AB$.

Let equation of $CD$ be $\;$ $y = mx + c$ $\;\;\; \cdots \; (2)$

Then slope of $CD = m = \dfrac{-1}{\text{slope of AB}} = \dfrac{1}{2}$

$\therefore \;$ Equation of $CD$ [equation $(2)$] becomes $\;\;$ $y = \dfrac{1}{2} x + c$ $\;\;\; \cdots \; (3)$

Since $CD$ passes through the point $C \left(2, 3\right)$, equation $(3)$ becomes

$3 = \dfrac{2}{2} + c$ $\implies$ $c = 2$

$\therefore \;$ Equation of $CD$ [equation $(3)$] is

$y = \dfrac{1}{2} x + 2$

i.e. $\;$ $x = 2y - 4$ $\;\;\; \cdots \; (4)$

Solving equations $(1)$ and $(4)$ simultaneously gives

$\left(2y - 4\right)^2 + y^2 - 4 \left(2y - 4\right) - 6y + 8 = 0$

i.e. $\;$ $4y^2 - 16 y + 16 + y^2 - 8y + 16 - 6y + 8 = 0$

i.e. $\;$ $5y^2 - 30y + 40 = 0$

i.e. $\;$ $y^2 - 6y + 8 = 0$

i.e. $\;$ $\left(y - 4\right) \left(y - 2\right) = 0$

i.e. $\;$ $y = 4$ $\;$ or $\;$ $y = 2$

$\therefore \;$ We have from equation $(4)$,

when $\;$ $y = 4$, $\;$ $x = 2 \times 4 - 4 = 4$

when $\;$ $y = 2$, $\;$ $x = 2 \times 2 - 4 = 0$

$\therefore \;$ The extremities of diameter $CD$ are $\left(4, 4\right)$ and $\left(0,2\right)$

Coordinate Geometry - Circle

The tangent at $\left(\alpha, \beta\right)$ to the circle $x^2 + y^2 = r^2$ cuts the axes of coordinates at $A$ and $B$. Prove that the area of $\triangle OAB = \dfrac{r^4}{2 \alpha \beta}$

Given point: $\;$ $\left(x_1, y_1\right) = \left(\alpha, \beta\right)$

Given circle: $\;$ $x^2 + y^2 = r^2$

Equation of tangent to the circle $\;$ $x^2 + y^2 = r^2$ $\;$ at the point $\;$ $\left(x_1, y_1\right)$ is $\;$ $x x_1 + yy_1 = r^2$

$\therefore \;$ Equation of tangent to the circle $\;$ $x^2 + y^2 = r^2$ $\;$ at the point $\;$ $\left(\alpha, \beta\right)$ $\;$ is

$\alpha x + \beta y = r^2$ $\;\;\; \cdots \; (1)$

Let equation $(1)$ cut the $X$ and the $Y$ axes at the points $A$ and $B$ respectively.

Then, $\;$ $A = \left(\dfrac{r^2}{\alpha}, 0\right)$, $\;$ $B = \left(0, \dfrac{r^2}{\beta}\right)$

$OA = \dfrac{r^2}{\alpha}$, $\;$ $OB = \dfrac{r^2}{\beta}$ $\;$ where $\;$ $O \left(0, 0\right)$ $\;$ is the origin

$\therefore \;$ Area of $\triangle OAB = \dfrac{1}{2} \times OA \times OB$

i.e. $\;$ Area of $\triangle OAB = \dfrac{1}{2} \times \dfrac{r^2}{\alpha} \times \dfrac{r^2}{\beta}$

i.e. $\;$ Area of $\triangle OAB = \dfrac{r^4}{2 \alpha \beta}$

Coordinate Geometry - Circle

$P$ is a point on the circle. A line is drawn through the center parallel to the tangent at $P$ and cutting the circle at $Q$. Prove that the tangent at $Q$ is perpendicular to the tangent at $P$.

Let $C$ be the center of the circle and $P$ any point on the circle.

$PA$ is the tangent at point $P$.

$CB$ is a line parallel to $PA$ and cutting the circle at the point $Q$.

$QD$ is the tangent to the circle at point $Q$.

The tangent to the circle at point $Q$ intersects $PA$ at the point $M$.

To prove that $\;$ $QD \perp PA$

i.e. $\;$ To prove that $\;$ $QM \perp PM$

From the figure,

$MP$ and $MQ$ are tangents drawn from the external point $M$.

$\therefore \;$ $MP = MQ$ $\;\;\; \cdots \; (1)$

$CP = CQ$ $\;\;\; \cdots \; (2)$ are the radius of the circle.

$\therefore \;$ $\angle CPM = 90^\circ$ $\;\;\; \cdots \; (3)$ $\;$ [radius is perpendicular to the tangent]

and $\;$ $\angle CQM = 90^\circ$ $\;\;\; \cdots \; (4)$

$\therefore \;$ From equations $(1), \; (2), \; (3) \;$ and $(4)$

$PMQC$ $\;$ is a square.

$\implies$ $\angle PMQ = 90^\circ$

i.e. $\;$ The tangent at $Q$ is perpendicular to the tangent at $P$.

Hence proved.

Coordinate Geometry - Circle

A circle with center at the origin, is described so as to touch the straight line $2y = x + 6$. Find its radius.

Let the equation of the required circle with center at the origin, i.e. $\;$ $\left(0, 0\right)$ $\;$ and radius $\;$ $r$ $\;$ be

$x^2 + y^2 = r^2$ $\;\;\; \cdots \; (1)$

Given line: $\;$ $2y = x + 6$

i.e. $\;$ $y = \dfrac{1}{2} x + 3$ $\;\;\; \cdots \; (2)$

Comparing equation $(2)$ with the standard equation of a straight line $\;$ $y = mx + c$ $\;$ gives

Slope $ = m = \dfrac{1}{2}$, $\;$ intercept on the $Y$ axis $= c = 3$

Equation $(2)$ will be a tangent to the circle given by equation $(1)$ if

$c^2 = \left(1 + m^2\right) r^2$

$\implies$ $r^2 = \dfrac{c^2}{1 + m^2}$

i.e. $\;$ $r^2 = \dfrac{3^2}{1 + \left(\dfrac{1}{2}\right)^2} = \dfrac{36}{5}$

$\therefore \;$ Radius of the circle $= r = \dfrac{6}{\sqrt{5}}$

Coordinate Geometry - Circle

Show that the circles $\;$ $x^2 + y^2 -8x - 4y + 10 = 0$ $\;$ and $\;$ $x^2 + y^2 - 10x - 10 y + 10 = 0$ $\;$ touch each other at the point $\left(3, -1\right)$. Also find the equation of the common tangent.

Equations of the given circles are

$x^2 + y^2 - 8x - 4y + 10 = 0$ $\;\;\; \cdots \; (1)$

$x^2 + y^2 - 10x - 10y + 10 = 0$ $\;\;\; \cdots \; (2)$

Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1 x + 2 f_1 y + c_1 = 0$ $\;$ gives

$g_1 = -4, \; f_1 = -2, \; c_1 = 10$

Center of circle [equation $(1)$] is $\;$ $C_1 = \left(-g_1, -f_1\right) = \left(4, 2\right)$

Radius of circle [equation $(1)$] is

$r_1 = \sqrt{g_1^2 + f_1^2 - c_1} = \sqrt{16 + 4 - 10} = \sqrt{10}$

Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_2 x + 2 f_2 y + c_2 = 0$ $\;$ gives

$g_2 = -5, \; f_2 = -5, \; c_2 = 10$

Center of circle [equation $(2)$] is $\;$ $C_2 = \left(-g_2, -f_2\right) = \left(5, 5\right)$

Radius of circle [equation $(2)$] is

$r_2 = \sqrt{g_2^2 + f_2^2 - c_2} = \sqrt{25 + 25 - 10} = \sqrt{40} = 2 \sqrt{10}$

Distance between the centers of the two circles

$= C_1 C_2 = \sqrt{\left(4 - 5\right)^2 + \left(2 - 5\right)^2} = \sqrt{1 + 9} = \sqrt{10}$

$\left|r_1 - r_2\right| = \left|\sqrt{10} - 2 \sqrt{10}\right| = \sqrt{10}$

i.e. $\;$ $C_1 C_2 = \left|r_1 - r_2\right|$

$\implies$ The two circles touch each other internally.

Subtracting equations $(1)$ and $(2)$ gives

$2x + 6y = 0$ $\implies$ $x = - 3y$ $\;\;\; \cdots \; (3)$

In view of equation $(3)$ equation $(1)$ becomes

$\left(-3y\right)^2 + y^2 - 8 \times \left(-3y\right) - 4y + 10 = 0$

i.e. $\;$ $9 y^2 + y^2 + 24 y - 4y + 10 = 0$

i.e. $\;$ $10 y^2 + 20 y + 10 = 0$

i.e. $\;$ $y^2 + 2y + 1 = 0$

i.e. $\;$ $\left(y + 1\right)^2 = 0$

$\implies$ $y = -1$

Substituting the value of $y$ in equation $(3)$ gives

$x = - 3 \times \left(-1\right) = 3$

$\therefore \;$ The point of contact of the two circles is $\left(x_1, y_1\right) = \left(3, -1\right)$

Equation of common tangent at the point $\left(x_1, y_1\right)$ to circle given by equation $(1)$ is

$xx_1 + yy_1 + g_1 \left(x + x_1\right) + f_1 \left(y + y_1\right) + c_1 = 0$

i.e. $\;$ $3x - y - 4 \left(x + 3\right) - 2 \left(y - 1\right) + 10 = 0$

i.e. $\;$ $3x - y - 12 - 2y + 2 + 10 = 0$

i.e. $\;$ $x + 3y = 0$

Coordinate Geometry - Circle

Show that the line $\;$ $3x = y + 13$ $\;$ touches the circle $\;$ $x^2 + y^2 - 4x - 6y + 3 = 0$ $\;$ and find the point of contact.

Given line: $\;\;$ $3x = y + 13$ $\;\;\; \cdots \; (1a)$

i.e. $\;$ $3x - y - 13 = 0$ $\;\;\; \cdots \; (1b)$

Given circle: $\;\;$ $x^2 + y^2 - 4x - 6y + 3 = 0$ $\;\;\; \cdots \; (2)$

Comparing equation $(2)$ with the standard equation of a circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$g = -2, \; f = -3, \; c = 3$

$\therefore \;$ Center of given circle $= C = \left(-g, -f\right) = \left(2, 3\right)$

Radius of given circle $= r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 - 3} = \sqrt{10}$

Perpendicular distance from the center $C \left(2, 3\right)$ of the circle to line [equation $(1b)$] is

$= d = \left|\dfrac{3 \times 2 - 1 \times 3 - 13}{\sqrt{3^2 + \left(-1\right)^2}}\right|$

i.e. $\;$ $d = \left|\dfrac{6 - 3 - 13}{\sqrt{10}}\right| = \dfrac{10}{\sqrt{10}} = \sqrt{10}$

$\implies$ Perpendicular distance from the center $C$ of the circle to the given line $=$ radius of given circle

i.e. $\;$ The given line touches the circle

Now, from equation $(1a)$, $\;$ $y = 3x - 13$

Substituting the value of $y$ in equation $(2)$ gives

$x^2 + \left(3x - 13\right)^2 - 4x - 6 \left(3x - 13\right) + 3 = 0$

i.e. $\;$ $x^2 + 9x^2 - 78x + 169 - 4x - 18x + 78 + 3 = 0$

i.e. $\;$ $10 x^2 - 100 x + 250 = 0$

i.e. $\;$ $x^2 - 10 x + 25 = 0$

i.e. $\;$ $\left(x - 5\right)^2 = 0$

i.e. $\;$ $x = 5$

Substituting the value of $x$ in equation $(1a)$ gives

$y = 3x - 13 = 3 \times 5 - 13 = 2$

$\therefore \;$ The point of contact is $\;$ $\left(x, y\right) = \left(5, 2\right)$

Coordinate Geometry - Circle

Show that the straight line $3x - 4y + 10 = 0$ touches the circle $x^2 + y^2 = 4$ and find the point of contact.

Given line: $\;\;$ $3x - 4y + 10 = 0$

i.e. $\;$ $y = \dfrac{3}{4} x + \dfrac{10}{4}$

i.e. $\;$ $y = \dfrac{3}{4} x + \dfrac{5}{2}$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of line $\;$ $y = mx + c$ $\;$ gives

$c = \dfrac{5}{2}, \;\; m = \dfrac{3}{4}$

Given circle: $\;\;$ $x^2 + y^2 = 4$ $\;\;\; \cdots \; (2)$

Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 = r^2$ $\;$ gives

$r^2 = 4$

Condition that the line $\;$ $y = mx + c$ $\;$ is a tangent to the circle $\;$ $x^2 + y^2 = r^2$ $\;$ is $\;\;$ $c^2 = r^2 \left(1 + m^2\right)$

Now,

$c^2 = \left(\dfrac{5}{2}\right)^2 = \dfrac{25}{4}$ $\;\;\; \cdots \; (3a)$

$r^2 \left(1 + m^2\right) = 4 \left(1 + \dfrac{9}{16}\right) = 4 \times \dfrac{25}{16} = \dfrac{25}{4}$ $\;\;\; \cdots \; (3b)$

$\therefore \;$ From equations $(3a)$ and $(3b)$, $\;\;$ $c^2 = r^2 \left(1 + m^2\right)$

$\implies$ The given line is a tangent to the given circle.

In view of equation $(1)$ equation $(2)$ becomes

$x^2 + \left(\dfrac{3}{4}x + \dfrac{5}{2}\right)^2 = 4$

i.e. $\;$ $x^2 + \dfrac{9}{16} x^2 + \dfrac{25}{4} + 2 \times \dfrac{3x}{4} \times \dfrac{5}{2} = 4$

i.e. $\;$ $\dfrac{25}{16} x^2 + \dfrac{15}{4} x + \dfrac{9}{4} = 0$

i.e. $\;$ $25 x^2 + 60 x + 36 = 0$

i.e. $\;$ $\left(5x + 6\right)^2 = 0$

$\implies$ $x = \dfrac{-6}{5}$

Substituting the value of $x$ in equation $(1)$ gives

$y = \dfrac{3}{4} \times \left(\dfrac{-6}{5}\right) + \dfrac{5}{2} = \dfrac{8}{5}$

$\therefore \;$ The point of contact is $\;\;$ $\left(\dfrac{-6}{5}, \dfrac{8}{5}\right)$

Coordinate Geometry - Circle

Find the equation of the line which is inclined at $45^\circ$ to the positive direction of the $X$ axis and on which the circles $x^2 + y^2 = 9$ and $x^2 + y^2 - 12x - 8y - 29 = 0$ cut equal intercepts.

Given equations of circles:

$x^2 + y^2 = 9$ $\;\;\; \cdots \; (1)$

$x^2 + y^2 - 12x - 8y - 29 = 0$ $\;\;\; \cdots \; (2)$

Let the equation of the required line be: $\;\;$ $y = mx + c$

Since the required line makes an angle of $45^\circ$ with the positive direction of the $X$ axis,

$\implies$ slope of the line $= m = \tan 45^\circ = 1$

$\therefore \;$ Equation of the required line is: $\;\;$ $y = x + c$ $\;\;\; \cdots \; (3)$

Let equations $(1)$ and $(3)$ intersect at the points $A$ and $B$.

We have, from equations $(1)$ and $(3)$

$x^2 + \left(x + c\right)^2 = 9$

i.e. $\;$ $2 x^2 + 2 c x + c^2 - 9 = 0$

$\implies$ $x = \dfrac{-2c \pm \sqrt{4 c^2 - 4 \times 2 \times \left(c^2 - 9\right)}}{2 \times 2}$

i.e. $\;$ $x = \dfrac{-2c \pm \sqrt{4 c^2 - 8c^2 + 72}}{4}$

i.e. $\;$ $x = \dfrac{-2c \pm \sqrt{72 - 4 c^2}}{4}$

i.e. $\;$ $\dfrac{-2 c \pm 2 \sqrt{18 - c^2}}{4}$

i.e. $\;$ $x = \dfrac{-c \pm \sqrt{18 - c^2}}{2}$

Substituting the value of $x$ in equation $(3)$ gives

$y = \dfrac{-c \pm \sqrt{18 - c^2}}{2} + c$

i.e. $y = \dfrac{c \pm \sqrt{18 - c^2}}{2}$

$\therefore \;$ The points of intersections are

$A = \left(\dfrac{-c + \sqrt{18 - c^2}}{2}, \dfrac{c + \sqrt{18 - c^2}}{2}\right)$, $\;$ $B = \left(\dfrac{-c - \sqrt{18 - c^2}}{2}, \dfrac{c - \sqrt{18 - c^2}}{2}\right)$

$AB$ is the length of the chord.

$\therefore \;$ $AB^2 = \left[\dfrac{-c - \sqrt{18 - c^2}}{2} - \left(\dfrac{c + \sqrt{18 - c^2}}{2}\right)\right]^2 + \left[\dfrac{c - \sqrt{18 - c^2}}{2} - \dfrac{c + \sqrt{18 - c^2}}{2}\right]^2$

i.e. $\;$ $AB^2 = 18 - c^2 + 18 - c^2$

i.e. $\;$ $AB^2 = 36 - 2c^2$ $\;\;\; \cdots \; (4)$

Let equations $(2)$ and $(3)$ intersect at the points $C$ and $D$.

Solving equations $(2)$ and $(3)$ simultaneously gives,

$x^2 + \left(x + c\right)^2 - 12 x - 8 \left(x + c\right) - 29 = 0$

i.e. $\;$ $x^2 + x^2 + 2cx + c^2 - 12x - 8x - 8c - 29 = 0$

i.e. $\;$ $2x^2 + \left(2c - 20\right)x + c^2 - 8c - 29 = 0$

i.e. $\;$ $x = \dfrac{20 - 2c \pm \sqrt{\left(2c - 20\right)^2 - 4 \times 2 \times \left(c^2 - 8c - 29\right)}}{2 \times 2}$

i.e. $\;$ $x = \dfrac{20 - 2c \pm \sqrt{4c^2 + 400 - 80c - 8c^2 + 64c + 232}}{4}$

i.e. $\;$ $x = \dfrac{20 - 2c \pm \sqrt{632 - 16c - 4c^2}}{4}$

i.e. $\;$ $x = \dfrac{10 - c \pm \sqrt{158 - 4c - c^2}}{2}$

Substituting the value of $x$ in equation $(3)$ gives,

$y = \dfrac{10 -c \pm \sqrt{158 - 4c - c^2}}{2} + c$

i.e. $y = \dfrac{10 + c \pm \sqrt{158 - 4c - c^2}}{2}$

$\therefore \;$ the points of intersection are

$C = \left(\dfrac{10 - c + \sqrt{158 - 4c - c^2}}{2}, \dfrac{10 + c + \sqrt{158 - 4c - c^2}}{2}\right)$ $\;$ and

$D = \left(\dfrac{10 - c - \sqrt{158 - 4c - c^2}}{2}, \dfrac{10 + c - \sqrt{158 - 4c - c^2}}{2}\right)$

$CD$ is the length of the chord.

$\therefore \;$ $CD^2 = \left(\dfrac{10 - c + \sqrt{158 - 4c - c^2}}{2} - \dfrac{10 - c - \sqrt{158 - 4c - c^2}}{2}\right)^2$
$\hspace{1.5cm}$ $+ \left(\dfrac{10 + c + \sqrt{158 - 4c - c^2}}{2} - \dfrac{10 + c - \sqrt{158 - 4c - c^2}}{2}\right)^2$

i.e. $\;$ $CD^2 = 158 - 4c - c^2 + 158 - 4c - c^2$

i.e. $\;$ $CD^2 = 316 - 8c - 2c^2$ $\;\;\; \cdots \; (5)$

Since the required line cuts equal intercepts on both the circles, we have

$AB = CD$

$\implies$ $AB^2 = CD^2$

i.e. $\;$ $36 - 2 c^2 = 316 - 8c - 2c^2$

i.e. $\;$ $8c = 280$ $\implies$ $c = 35$

$\therefore \;$ The required equation of line is

$y = x + 35$

Coordinate Geometry - Circle

Find the equation to the circle which has its center at $\left(1, -2\right)$ and touches the line $7x - 24y - 5 = 0$.

Center of the required circle $= \left(h, k\right) = \left(1, -2\right)$

The required circle touches the line $\;$ $7x - 24y -5 = 0$

$\therefore \;$ Perpendicular distance from the center $\left(0, 0\right)$ to the point where given line touches the circle is the radius of the required circle.

i.e. $\;$ $r = \left|\dfrac{7 \times 1 - 24 \times \left(-2\right) - 5}{\sqrt{7^2 + \left(-24\right)^2}}\right|$

i.e. $\;$ $r = \left|\dfrac{7 + 48 - 5}{\sqrt{49 + 576}}\right| = \left|\dfrac{50}{\sqrt{625}}\right|$

i.e. $\;$ $r = \dfrac{50}{25} = 2$

Equation of circle with center at $\left(h, k\right)$ and radius $= r$ is $\;$ $\left(x - h\right)^2 + \left(y - k\right)^2 = r^2$

$\therefore \;$ Equation of required circle is

$\left(x - 1\right)^2 + \left(y - 2\right)^2 = 2^2$

i.e. $\;$ $x^2 + y^2 - 2x + 4y + 1 = 0$

Coordinate Geometry - Circle

Find the equation to the circle which has its center at $\left(0, 0\right)$ and touches the line $12x + 5y - 11 = 0$.

Center of the required circle $= \left(0, 0\right)$

The required circle touches the line $\;$ $12x + 5y - 11 = 0$

$\therefore \;$ Perpendicular distance from the center $\left(0, 0\right)$ to the point where given line touches the circle is the radius of the required circle.

i.e. $\;$ $r = \left|\dfrac{12 \times 0 + 5 \times 0 - 11}{\sqrt{12^2 + 5^2}}\right|$

i.e. $\;$ $r = \dfrac{11}{\sqrt{144 + 25}} = \dfrac{11}{13}$

Equation of circle with center at $\left(0, 0\right)$ and radius $= r$ is $\;$ $x^2 + y^2 = r^2$

$\therefore \;$ Equation of required circle is

$x^2 + y^2 = \left(\dfrac{11}{13}\right)^2$

i.e. $\;$ $169 x^2 + 169 y^2 = 121$

Coordinate Geometry - Circle

Find the tangents to the circle $x^2 + y^2 + 2x + 2y = 7$ inclined at $45^\circ$ to the positive direction of the $X$ axis.

Equation of given circle: $\;\;\;$ $x^2 + y^2 + 2x + 2y = 7$

i.e. $\;$ $x^2 + y^2 + 2x + 2y - 7 = 0$ $\;\;\; \cdots \; (1)$

Comparing with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$g = 1, \; f = 1, \; c = -7$

Radius of the circle $= r = \sqrt{g^2 + f^2 - c} = \sqrt{1 + 1 + 7} = 3$

Inclination of the required tangent to the positive direction of $X$ axis $= \theta = 45^\circ$

$\therefore \;$ Slope of tangent $= m = \tan 45^\circ = 1$

Let the equation of the required tangent be $y = mx + c_1$

Condition that the line $\;$ $y = mx + c_1$ $\;$ is a tangent to the circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ is

$c_1 = \pm r \sqrt{1 + m^2}$

i.e. $\;$ $c_1 = \pm 3 \sqrt{1 + 1^2} = \pm 3 \sqrt{2}$

$\therefore \;$ The equations of the tangents are

$y = x + 3 \sqrt{2}$ $\;$ and $\;$ $y = x - 3 \sqrt{2}$

Coordinate Geometry - Circle

Find the tangents to the circle $x^2 + y^2 + 8x - 4y = 5$ parallel to the line $2x + 3y + 5 = 0$.

Equation of given circle: $\;\;\;$ $x^2 + y^2 + 8x - 4y = 5$

i.e. $\;$ $x^2 + y^2 + 8x - 4y - 5 = 0$ $\;\;\; \cdots \; (1)$

Comparing with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$g = 4, \; f = -2, \; c = -5$

Radius of the circle $= r = \sqrt{g^2 + f^2 - c} = \sqrt{16 + 4 + 5} = 5$

Equation of straight line: $\;\;\;$ $2x + 3y + 5 = 0$

Slope of the given straight line $= m = \dfrac{-2}{3}$

Any line parallel to the given straight line is $\;$ $2x + 3y + c_1 = 0$ $\;\;\; \cdots \; (2)$

Slope of line parallel to the given line $= m = \dfrac{-2}{3}$

Condition that the line $\;$ $y = mx + c_1$ $\;$ is a tangent to the circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ is

$c_1 = \pm r \sqrt{1 + m^2}$

i.e. $\;$ $c_1 = \pm 5 \sqrt{1 + \dfrac{4}{9}} = \pm \dfrac{5 \sqrt{13}}{3}$

$\therefore \;$ The equations of tangents are

$2x + 3y \pm \dfrac{5 \sqrt{13}}{3} = 0$

i.e. $\;$ $6x + 9y \pm 5 \sqrt{13} = 0$

Coordinate Geometry - Circle

Find the tangents to the circle $x^2 + y^2 = 81$ parallel to the line $3x + 4y = 13$.

Equation of given circle: $\;\;\;$ $x^2 + y^2 = 81$ $\;\;\; \cdots \; (1)$

Comparing with the standard equation of circle $\;$ $x^2 + y^2 = r^2$ $\;$ gives

$r^2 = 81$

Equation of straight line: $\;\;\;$ $3x + 4y = 13$

Any line parallel to the given straight line is $\;$ $3x + 4y + c = 0$ $\;\;\; \cdots \; (2)$

Equation $(2)$ will touch the given circle, whose center is $= \left(0, 0\right)$ and radius is $= 9$ if

$\left|\dfrac{3 \times 0 + 4 \times 0 + c}{\sqrt{3^2 + 4^2}}\right| = 9$

i.e. $\;$ $\dfrac{\pm c}{5} = 9$

i.e. $\;$ $c = \pm 45$

$\therefore \;$ The equations of tangents are

$3x + 4y + 45 = 0$ $\;\;$ and $\;\;$ $3x + 4y - 45 = 0$

OR

Condition that the line $\;$ $y = mx + c$ $\;$ is a tangent to the circle $\;$ $x^2 + y^2 = r^2$ $\;$ is

$c = \pm r \sqrt{1 + m^2}$

i.e. $\;$ $c = \pm 9 \sqrt{1 + \left(\dfrac{-3}{4}\right)^2}$

i.e. $\;$ $c = \pm 9 \sqrt{\dfrac{25}{16}}$

i.e. $\;$ $c = \pm \dfrac{45}{4}$

$\therefore \;$ The equations of tangents are

$y = \dfrac{-3}{4}x \pm \dfrac{45}{4}$

i.e. $\;$ $3x + 4y \pm 45 = 0$

Coordinate Geometry - Circle

Find the condition that the straight line $x \cos \alpha + y \sin \alpha = p$ may touch the circle $x^2 + y^2 = r^2$.

Equation of given circle: $\;\;\;$ $x^2 + y^2 = r^2$ $\;\;\; \cdots \; (1)$

Equation of straight line: $\;\;\;$ $x \cos \alpha + y \sin \alpha = p$ $\;\;\; \cdots \; (2)$

i.e. $\;$ $y = \dfrac{p - x \cos \alpha}{\sin \alpha}$

The $x$ coordinates of the points of intersection of equations $(1)$ and $(2)$ are given by the equation

$x^2 + \left(\dfrac{p - x \cos \alpha}{\sin \alpha}\right)^2 = r^2$

i.e. $\;$ $x^2 \sin^2 \alpha + p^2 - 2px \cos \alpha + x^2 \cos^2 \alpha = r^2 \sin^2 \alpha$

i.e. $\;$ $x^2 - 2p x \cos \alpha + p^2 - r^2 \sin^2 \alpha = 0$ $\;\;\; \cdots \; (3)$

Equation $(2)$ will be a tangent to $(1)$ if the quadratic given by $(3)$ has equal roots.

i.e. $\;$ if the discriminant of quadratic equation $(3)$ is zero.

i.e. if $\;$ $4 p^2 \cos^2 \alpha - 4 \left(p^2 - r^2 \sin^2 \alpha\right) = 0 $

i.e. $\;$ $p^2 \left(1 - \cos^2 \alpha\right) = r^2 \sin^2 \alpha$

i.e. $\;$ $p^2 \sin^2 \alpha = r^2 \sin^2 \alpha$

i.e. $\;$ $p = \pm r$

Coordinate Geometry - Circle

Find the points of intersection of the circles $x^2 + y^2 = 130$ and $x^2 + y^2 - 10x - 2y - 26 = 0$ and find the tangents at their common points.

Equations of given circles:

$x^2 + y^2 = 130$ $\;\;\; \cdots \; (1)$

$x^2 + y^2 - 10x - 2y - 26 = 0$ $\;\;\; \cdots \; (2)$

In view of equation $(1)$, equation $(2)$ can be written as

$130 - 10x - 2y - 26 = 0$

i.e. $\;$ $10x + 2y = 104$

$\implies$ $y = 52 - 5x$ $\;\;\; \cdots \; (3)$

In view of equation $(3)$ equation $(1)$ becomes

$x^2 + \left(52 - 5x\right)^2 = 130$

i.e. $\;$ $x^2 + 2704 - 520 x + 25x^2 = 130$

i.e. $\;$ $26x^2 - 520x + 2574 = 0$

i.e. $\;$ $x^2 - 20 x + 99 = 0$

$\implies$ $x = 11$ $\;$ or $\;$ $x = 9$

We have from equation $(3)$,

when $\;$ $x = 11$, $\;$ $y = 52 - 5 \times 11 = -3$

when $\;$ $x = 9$, $\;$ $y = 52 - 5 \times 9 = 7$

$\therefore \;$ The points of intersections of the two given circles are

$\left(11, -3\right)$ $\;$ and $\;$ $\left(9, 7\right)$

Equation of tangent to the circle $x^2 + y^2 = r^2$ at any given point $\left(x_1, y_1\right)$ is

$xx_1 + yy_1 = r^2$

$\therefore \;$ Equation of tangent to circle given by equation $(1)$ at $\left(11, -3\right)$ is

$11x - 3y = 130$

Equation of tangent to circle given by equation $(1)$ at $\left(9, 7\right)$ is

$9x + 7y = 130$

Coordinate Geometry - Circle

Show that the point $\left(-9, -1\right)$ lies on the circle $x^2 + y^2 + 4x + 8y - 38 = 0$.
Find the point diametrically opposite to it and find the tangents at these points.

Equation of given circle: $\;$ $x^2 + y^2 + 4x + 8y - 38 = 0$ $\;\;\; \cdots \; (1)$

Given point: $\;$ $P \left(x_1, y_1\right) = \left(-9, -1\right)$

Substituting the given point in the equation of the circle gives

$\left(-9\right)^2 + \left(-1\right)^2 + 4 \times \left(-9\right) + 8 \times \left(-1\right) - 38 = 0$

i.e. $\;$ $81 + 1 - 36 - 8 - 38 = 0$

i.e. $\;$ $0 = 0$ $\;\;$ which is true.

$\implies$ The point $\left(-9, -1\right)$ lies on the given circle.

Comparing the given equation of circle with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$g = 2$, $\;$ $f = 4$, $\;$ $c = -38$

Radius of the given circle $= r = \sqrt{g^2 + f^2 - c}$

i.e. $\;$ $r = \sqrt{2^2 + 4^2 - \left(-38\right)} = \sqrt{4 + 16 + 38} = \sqrt{58}$

$\therefore \;$ Diameter of the circle $= d = 2r = 2 \sqrt{58}$

Let $Q \left(x_2, y_2\right)$ be the point diametrically opposite to $P$.

Then, $PQ$ is the diameter of the circle.

i.e. $\;$ $PQ = \sqrt{\left(x_2 + 9\right)^2 + \left(y_2 + 1\right)^2} = 2\sqrt{58}$

i.e. $\;$ $x_2^2 + 18x_2 + 81 + y_2^2 + 2y_2 + 1 = 232$

i.e. $\;$ $x_2^2 + y_2^2 + 18x_2 + 2y_2 - 150 = 0$ $\;\;\; \cdots \; (2)$

$Q \left(x_2, y_2\right)$ also lies on equation $(1)$.

$\therefore \;$ We have,

$x_2^2 + y_2^2 + 4x_2 + 8y_2 - 38 = 0$ $\;\;\; \cdots \; (3)$

Subtracting equations $(2)$ and $(3)$ gives

$14x_2 - 6y_2 - 112 = 0$

i.e. $\;$ $7x_2 - 3y_2 - 56 = 0$ $\;\;\; \cdots \; (4)$

$\implies$ $x_2 = \dfrac{3y_2 + 56}{7}$

Substituting the value of $x_2$ in equation $(3)$ gives

$\left(\dfrac{3y_2 + 56}{7}\right)^2 + y_2^2 + 4 \times \left(\dfrac{3y_2 + 56}{7}\right) + 8y_2 - 38 = 0$

i.e. $\;$ $\dfrac{9y_2^2 + 336 y_2 + 3136}{49} + y_2^2 + \dfrac{12y_2 + 224}{7} + 8y_2 - 38 = 0$

i.e. $\;$ $9y_2^2 + 336 y_2 + 3136 + 49 y_2^2 + 84 y_2 + 1568 + 392 y_2 - 1862 = 0$

i.e. $\;$ $58 y_2^2 + 812 y_2 + 2842 = 0$

i.e. $\;$ $y_2^2 + 14y_2 + 49 = 0$

i.e. $\;$ $\left(y_2 + 7\right)^2 = 0$ $\implies$ $y_2 = -7$

Substituting the value of $y_2$ in equation $(4)$ gives

$7 x_2 - 3 \times \left(-7\right) - 56 = 0$

i.e. $\;$ $7 x_2 - 35 = 0$ $\implies$ $x_2 = 5$

$\therefore \;$ $Q \left(x_2, y_2\right) = \left(5, -7\right)$

Equation of tangent to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ at any given point $\left(x_1, y_1\right)$ is

$xx_1 + yy_1 + g \left(x + x_1\right) + f \left(y + y_1\right) + c = 0$

$\therefore \;$ Equation of tangent at $\left(-9, -1\right)$ is

$-9x - y + 2 \left(x - 9\right) + 4 \left(y - 1\right) - 38 = 0$

i.e. $\;$ $-9x - y + 2x -18 + 4y - 4 - 38 = 0$

i.e. $\;$ $7x - 3y + 60 = 0$

Equation of tangent at $\left(5, -7\right)$ is

$5x - 7y + 2 \left(x + 5\right) + 4 \left(y - 7\right) - 38 = 0$

i.e. $\;$ $5x - 7y + 2x + 10 + 4y - 28 - 38 = 0$

i.e. $\;$ $7x - 3y - 56 = 0$

Coordinate Geometry - Circle

Find the equation of the tangent to the circle $x^2 + y^2 + 8x + 10y - 89 = 0$ at the point $\left(-1, 6\right)$.

Equation of given circle: $\;$ $x^2 + y^2 + 8x + 10y - 89 = 0$

Comparing with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$g = 4$, $\;$ $f = 5$, $\;$ $c = -89$

Given point: $\;$ $\left(x_1, y_1\right) = \left(-1, 6\right)$

Equation of tangent to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ at any given point $\left(x_1, y_1\right)$ is

$xx_1 + yy_1 + g \left(x + x_1\right) + f \left(y + y_1\right) + c = 0$

$\therefore \;$ Required equation of tangent is

$-x + 6y + 4 \left(x - 1\right) + 5 \left(y + 6\right) - 89 = 0$

i.e. $\;$ $-x + 6y + 4x -4 + 5y + 30 - 89 = 0$

i.e. $\;$ $3x + 11y - 63 = 0$

Coordinate Geometry - Circle

Find the equation of the tangent to the circle $x^2 + y^2 = 85$ at the point $\left(7, 6\right)$.

Equation of given circle: $\;$ $x^2 + y^2 = 85$

Comparing with the standard equation of circle $\;$ $x^2 + y^2 = r^2$ $\;$ gives

$r^2 = 85$

Given point: $\;$ $\left(x_1, y_1\right) = \left(7, 6\right)$

Equation of tangent to the circle $x^2 + y^2 = r^2$ at any given point $\left(x_1, y_1\right)$ is

$xx_1 + yy_1 = r^2$

$\therefore \;$ Required equation of tangent is

$7x + 6y = 85$

Coordinate Geometry - Circle

Find $\ell$ so that the point dividing the joint of $\left(4,7\right)$ and $\left(-5,6\right)$ in the ratio $\ell : 1$ should lie on the circle $x^2 + y^2 = 65$.

Let the given points be $A \left(4,7\right)$ and $B \left(-5, 6\right)$

Let $P \left(p, q\right)$ be the point dividing the join of points $A$ and $B$ in the ratio $\ell : 1$

Then, by section formula for internal division

$p = \dfrac{-5 \ell + 4}{\ell + 1}$ $\;$ and $\;$ $q = \dfrac{6 \ell + 7}{\ell + 1}$

Equation of the given circle is $\;$ $x^2 + y^2 = 65$

Since $P \left(p, q\right)$ lies on the given circle

$p^2 + q^2 = 65$

i.e. $\;$ $\left(\dfrac{-5 \ell + 4}{\ell + 1}\right)^2 + \left(\dfrac{6 \ell + 7}{\ell + 1}\right)^2 = 65$

i.e. $\;$ $25 \ell ^2 - 40 \ell + 16 + 36 \ell ^ 2 + 84 \ell + 49 = 65 \ell ^2 + 130 \ell + 65$

i.e. $\;$ $4 \ell ^2 + 86 \ell = 0$

i.e. $\;$ $2 \ell ^2 + 43 \ell = 0$

i.e. $\;$ $\ell \left(2 \ell + 43\right) = 0$

i.e. $\;$ $\ell = 0$ $\;$ or $2 \ell + 43 = 0$

$\because \;$ $\ell = 0$ is not possible for the given internal division

$\implies$ $2 \ell + 43 = 0$

$\implies$ $\ell = \dfrac{-43}{2}$