Coordinate Geometry - Circle

Find the equation to the circle which passes through the points $\left(-1, 0\right)$, $\left(3, 4\right)$ and has its center on the line $\;$ $3x - y - 23 = 0$

Let the equation of the required circle be $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (1)$

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

Given: Center lies on the line $\;$ $3x - y - 23 = 0$

$\therefore \;$ We have $\;$ $- 3g + f - 23 = 0$ $\;\;\; \cdots \; (2)$

The required circle passes through the points $\;$ $\left(-1, 0\right)$ $\;$ and $\;$ $\left(3, 4\right)$

$\therefore \;$ We have from equation $(1)$,

$\left(-1\right)^2 = 0^2 + 2 \times g \times \left(-1\right) + 2 \times f \times 0 + c = 0$

i.e. $\;$ $- 2 g + c + 1 = 0$

$\implies$ $g = \dfrac{c + 1}{2}$ $\;\;\; \cdots \; (3)$

and $\;$ $3^2 + 4^2 + 2 \times g \times 3 + 2 \times f \times 4 + c = 0$

i.e. $\;$ $9 + 16 + 6g + 8f + c = 0$

i.e. $\;$ $6g + 8f + c + 25 = 0$ $\;\;\; \cdots \; (4)$

Substituting the value of $g$ from equation $(3)$ in equation $(4)$ gives

$6 \left(\dfrac{c + 1}{2}\right) + 8 f + c + 25 = 0$

i.e. $\;$ $3c + 3 + 8f + c + 25 = 0$

i.e. $\;$ $8f + 4c + 25 = 0$

$\implies$ $f = \dfrac{-c - 7}{2}$ $\;\;\; \cdots \; (5)$

In view of equations $(4)$ and $(5)$, equation $(2)$ becomes,

$-3 \left(\dfrac{c + 1}{2}\right) - \left(\dfrac{c + 7}{2}\right) - 23 = 0$

i.e. $\;$ $\dfrac{-3c}{2} - \dfrac{3}{2} - \dfrac{c}{2} - 23 = 0$

i.e. $\;$ $-2c - 5 - 23 = 0$

i.e. $\;$ $2c = -28$ $\implies$ $c = -14$

Substituting the value of $c$ in equations $(3)$ and $(5)$ gives

$g = \dfrac{-14 + 1}{2} = \dfrac{-13}{2}$ $\;$ and $\;$ $f = \dfrac{14 - 7}{2} = \dfrac{7}{2}$

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

$x^2 + y^2 + 2 \times \left(\dfrac{-13}{2}\right) \times x +2 \times \dfrac{7}{2} \times y - 14 = 0$

i.e. $\;$ $x^2 + y^2 - 13x + 7y -14 = 0$

Coordinate Geometry - Circle

Find the length of the intercept made on the $Y$ axis by the circle $\;$ $x^2 + y^2 -2ax \sec \theta - 2by \tan \theta - b^2 = 0$

Equation of given circle is $\;\;$ $x^2 + y^2 -2ax \sec \theta - 2by \tan \theta - b^2 = 0$

When the circle cuts the $Y$ axis, its $x$ coordinate equals $0$.

Putting $x = 0$ in the equation of the given circle, we have,

$y^2 - 2by \tan \theta - b^2 = 0$

Solving for $y$, we get,

$\begin{aligned} y & = \dfrac{2b \tan \theta \pm \sqrt{\left(-2b \tan \theta\right)^2 - 4 \times 1 \times \left(-b^2\right)}}{2} \\\\ & = \dfrac{2 b \tan \theta \pm \sqrt{4b^2 \tan^2 \theta + 4b^2}}{2} \\\\ & = \dfrac{2b \tan \theta \pm \sqrt{4b^2 \left(1 + \tan^2 \theta\right)}}{2} \\\\ & = \dfrac{2 b \tan \theta \pm 2 b \sec \theta}{2} \\\\ & = b \tan \theta \pm b \sec \theta \end{aligned}$

$\therefore \;$ The given circle cuts the $Y$ axis at the points $\;$ $A \left(0, b \tan \theta + b \sec \theta\right)$ $\;$ and $\;$ $B \left(0, b \tan \theta - b \sec \theta\right)$

$\therefore \;$ Length of intercept made on the $Y$ axis by the given circle is

$= AB = \sqrt{\left(b \tan \theta + b \sec \theta - b \tan \theta + b \sec \theta\right)^2}$

i.e. $\;$ $AB = 2b \sec \theta$

Coordinate Geometry - Circle

Find the equation of the circle whose center is $\left(6, 8\right)$ and which passes through the center of the circle $x^2 + y^2 + 12x + 8y = 48$

Equation of given circle is

$x^2 + y^2 + 12x + 8y = 48$ $\;$ i.e. $\;$ $x^2 + y^2 + 12x + 8y - 48 = 0$

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

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

Center of the given circle $= \left(-g, -f\right) = \left(-6, -4\right)$

The required circle has center at $\left(6, 8\right)$ and passes through the point $\left(-6, -4\right)$

$\therefore \;$ Radius of the required circle $\;$ $= \sqrt{\left(6 + 6\right)^2 + \left(8 + 4\right)^2} = \sqrt{288}$

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

$\left(x - 6\right)^2 + \left(y - 8\right)^2 = \left(\sqrt{288}\right)^2$

i.e. $\;$ $x^2 -12x + 36 + y^2 - 16y + 64 = 288$

i.e. $\;$ $x^2 + y^2 -12x - 16y - 188 = 0$

Coordinate Geometry - Circle

Find the locus of a point which moves such that the sum of the squares of its distances from $\left(0, b\right)$, $\left(0, -b\right)$ is $2r^2$.

Let $P \left(x, y\right)$ be a point on the required locus.

Distance of $P \left(x, y\right)$ from $\left(0, b\right)$ is $= \sqrt{x^2 + \left(y - b\right)^2}$

Distance of $P \left(x, y\right)$ from $\left(0, -b\right)$ is $= \sqrt{x^2 + \left(y + b\right)^2}$

As per the given problem,

$\left(\sqrt{x^2 + \left(y - b\right)^2}\right)^2 + \left(\sqrt{x^2 + \left(y + b\right)^2}\right)^2 = 2r^2$

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

i.e. $\;$ $2x^2 + y^2 + b^2 - 2by + y^2 + b^2 + 2by = 2r^2$

i.e. $\;$ $2x^2 + 2y^2 + 2b^2 = 2r^2$

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

$x^2 + y^2 = r^2 - b^2$

Coordinate Geometry - Circle

Find the circle which passes through $\left(h, k\right)$ and whose center is $\left(a, b\right)$.

Center of the required circle $= \left(a, b\right)$.

The required circle passes through the point $\left(h, k\right)$.

$\therefore \;$ Radius of the required circle $= \sqrt{\left(a - h\right)^2 + \left(b - k\right)^2}$

The equation of the required circle is

$\left(x - a\right)^2 + \left(y - b\right)^2 = \left(\sqrt{\left(a - h\right)^2 + \left(b - k\right)^2}\right)^2$

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

i.e. $\;$ $x^2 + a^2 -2ax + y^2 + b^2 - 2by = a^2 + h^2 - 2ah + b^2 + k^2 - 2bk$

i.e. $\;$ $x^2 + y^2 - 2ax - 2by - h^2 - k^2 + 2ah + 2bk = 0$

Coordinate Geometry - Circle

Find the locus of a point which moves such that $k$ times its distance from the $Y$ axis is equal to the square of its distance from the origin.

Find the center and the radius of the resulting locus.

Let $P \left(x, y\right)$ be a point on the required locus.

Distance of point $P$ from the $Y$ axis is $= x$

Distance of point $P$ from the origin $= \sqrt{\left(x - 0\right)^2 + \left(y - 0\right)^2} = \sqrt{x^2 + y^2}$

As per question,

$k x = \left(\sqrt{x^2 + y^2}\right)^2$

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

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

is the equation of the required locus which is the equation of a circle.

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

$2g = -k$ $\implies$ $g = \dfrac{-k}{2}$, $\;$ $f = 0$, $\;$ $c = 0$

Center of the circle $= \left(-g, -f\right) = \left(\dfrac{k}{2}, 0\right)$

Radius of the circle $= \sqrt{g^2 + f^2 - c} = \sqrt{\left(\dfrac{-k}{2}\right)^2 + 0 + 0} = \sqrt{\dfrac{k^2}{4}} = \dfrac{k}{2}$

Coordinate Geometry - Circle

Find the equation of the circle which passes through the points $\left(h, k\right)$, $\left(k, k\right)$ and $\left(k, h\right)$.

Let the equation of the required circle be $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$

Since the required circle passes through the points $\left(h, k\right)$, $\left(k, k\right)$ and $\left(k, h\right)$, we have

$h^2 + k^2 + 2hg + 2kf + c = 0$ $\;\;\; \cdots \; (1)$

$k^2 + k^2 + 2kg + 2kf + c = 0$ $\;\;$ i.e. $\;\;$ $2k^2 + 2kg + 2kf + c = 0$ $\;\;\; \cdots \; (2)$

$k^2 + h^2 + 2kg + 2hf + c = 0$ $\;\;\; \cdots \; (3)$

Subtracting equation $(1)$ from equation $(2)$ we get,

$-h^2 + k^2 + 2g \left(k - h\right) = 0$ $\;\;\; \cdots \; (4)$

Subtracting equation $(3)$ from equation $(2)$ we get,

$-h^2 + k^2 + 2f \left(k - h\right) = 0$ $\;\;\; \cdots \; (5)$

We have from equations $(4)$ and $(5)$,

$2 \left(k - h\right) \left(g - f\right) = 0$ $\;\;\; \cdots \; (6)$

If $\;$ $k - h = 0$ $\;$ in equation $(6)$, then the coordinates become trivial $\implies$ $k - h \neq 0$

$\therefore \;$ In equation $(6)$, $\;\;\;$ $g - f = 0$ $\implies$ $g = f$

Substituting $g = f$ in equation $(1)$ gives,

$h^2 + k^2 + 2hf + 2kf + c = 0$ $\;\;\; \cdots \; (7)$

Substituting $g = f$ in equation $(2)$ gives,

$2k^2 + 2kf + 2kf +c = 0$

i.e. $\;$ $2k^2 + 4kf + c = 0$

$\implies$ $c = - 2k^2 - 4kf$ $\;\;\; \cdots \; (8)$

Substituting the value of $c$ from equation $(8)$ in equation $(7)$ gives

$h^2 + k^2 + 2hf + 2kf - 2k^2 -4kf = 0$

i.e. $\;$ $h^2 - k^2 + 2hf - 2 kf = 0$

i.e. $\;$ $h^2 - k^2 + 2f \left(h - k\right) = 0$

i.e. $\;$ $\left(h + k\right) \left(h - k\right) + 2 f \left(h - k\right) = 0$

i.e. $\;$ $h + k + 2f = 0$ $\;\;\;$ $\left[\because \; k - h \neq 0\right]$

$\implies$ $f = \dfrac{- \left(h + k\right)}{2}$

$\therefore \;$ $g = f = \dfrac{- \left(h + k\right)}{2}$

Substituting the value of $f$ in equation $(8)$ gives

$c = - 2k^2 - 4 \times \left[\dfrac{- \left(h + k\right)}{2}\right]k$

i.e. $\;$ $c = - 2k^2 + 2hk + 2k^2$ $\implies$ $c = 2hk$

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

$x^2 + y^2 + 2 \times \left[\dfrac{- \left(h + k\right)}{2}\right] x + 2 \times \left[\dfrac{- \left(h + k\right)}{2}\right]y + 2hk = 0$

i.e. $\;$ $x^2 + - \left(h + k\right)x - \left(h + k\right)y + 2hk = 0$

Coordinate Geometry - Circle

Find the equation of the circle which passes through the points $\left(\dfrac{1}{3}, \dfrac{2}{3}\right)$, $\left(\dfrac{-1}{3}, \dfrac{2}{3}\right)$ and $\left(\dfrac{4}{3}, \dfrac{5}{3}\right)$.

Let the equation of the required circle be $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$

Since the required circle passes through the points $\left(\dfrac{1}{3}, \dfrac{2}{3}\right)$, $\left(\dfrac{-1}{3}, \dfrac{2}{3}\right)$ and $\left(\dfrac{4}{3}, \dfrac{5}{3}\right)$, we have

$\left(\dfrac{1}{3}\right)^2 + \left(\dfrac{2}{3}\right)^2 + 2 \times \dfrac{1}{3} \times g + 2 \times \dfrac{2}{3} \times f + c = 0$

i.e. $\;$ $\dfrac{1}{9} + \dfrac{4}{9} + \dfrac{2}{3}g + \dfrac{4}{3}f + c = 0$

i.e. $\;$ $5 + 6g + 12f + 9c = 0$ $\;\;\; \cdots \; (1)$

$\left(\dfrac{-1}{3}\right)^2 + \left(\dfrac{2}{3}\right)^2 + 2 \times \dfrac{-1}{3} \times g + 2 \times \dfrac{2}{3} \times f + c = 0$

i.e. $\;$ $\dfrac{1}{9} + \dfrac{4}{9} - \dfrac{2}{3}g + \dfrac{4}{3}f + c = 0$

i.e. $\;$ $5 - 6g + 12f + 9c = 0$ $\;\;\; \cdots \; (2)$

$\left(\dfrac{4}{3}\right)^2 + \left(\dfrac{5}{3}\right)^2 + 2 \times \dfrac{4}{3} \times g + 2 \times \dfrac{5}{3} \times f + c = 0$

i.e. $\;$ $\dfrac{16}{9} + \dfrac{25}{9} + \dfrac{8}{3}g + \dfrac{10}{3}f + c = 0$

i.e. $\;$ $41 + 24g + 30f + 9c = 0$ $\;\;\; \cdots \; (3)$

Subtracting equation $(1)$ from equation $(2)$ we have, $\;\;\;$ $g = 0$

Substituting $g = 0$ in equations $(1)$ and $(3)$ we have,

$5 + 12f + 9c = 0$ $\;\;\; \cdots \; (4)$

$41 + 30f + 9c = 0$ $\;\;\; \cdots \; (5)$

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

$f = -2$ $\;\;$ and $\;$ $c = \dfrac{19}{9}$

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

$x^2 + y^2 + 2 \times 0 \; x + 2 \times \left(-2\right)y + \dfrac{19}{9} = 0$

i.e. $\;$ $9 x^2 + 9 y^2 - 36y + 19 = 0$

Coordinate Geometry - Circle

Find the equation of the circle which passes through the points $\left(6,1\right)$, $\left(4,3\right)$ and $\left(7,9\right)$.

Let the equation of the required circle be $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$

Since the required circle passes through the points $\left(6,1\right)$, $\left(4,3\right)$ and $\left(7,9\right)$, we have

$6^2 + 1^2 + 2 \times 6 \times g + 2 \times 1 \times f + c = 0$ $\;\;$ i.e. $\;$ $37 + 12g + 2f + c = 0$ $\;\;\; \cdots \; (1)$

$4^2 + 3^2 + 2 \times 4 \times g + 2 \times 3 \times f + c = 0$ $\;\;$ i.e. $\;$ $25 + 8 g + 6f + c = 0$ $\;\;\; \cdots \; (2)$

$7^2 + 9^2 + 2 \times 7 \times g + 2 \times 9 \times f + c = 0$ $\;\;$ i.e. $\;$ $130 + 14 g + 18 f + c = 0$ $\;\;\; \cdots \; (3)$

Subtracting equation $(1)$ from equation $(2)$ we have,

$-4g + 4f - 12 = 0$ $\;\;$ i.e. $\;$ $g - f = -3$ $\;\;\; \cdots \; (4)$

Subtracting equation $(2)$ from equation $(3)$ gives

$6g + 12f + 105 = 0$ $\;\;$ i.e. $\;$ $2g + 4f = -35$ $\;\;\; \cdots \; (5)$

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

$g = \dfrac{-47}{6}$ $\;\;$ and $\;$ $f = \dfrac{-29}{6}$

Substituting the values of $f$ and $g$ in equation $(1)$ gives

$37 + 12 \times \left(\dfrac{-47}{6}\right) + 2 \times \left(\dfrac{-29}{6}\right) + c = 0$ $\implies$ $c = \dfrac{200}{3}$

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

$x^2 + y^2 + 2 \times \left(\dfrac{-47}{6}\right) x + 2 \times \left(\dfrac{-29}{6}\right)y + \dfrac{200}{3} = 0$

i.e. $\;$ $3 x^2 + 3 y^2 - 47x - 29y + 200 = 0$

Coordinate Geometry - Circle

Determine the nature of the loci $\;$ $\left(x + 7\right) \left(x + 9\right) + \left(y - 1\right) \left(y + 3\right) = 11$

Equation of given locus is: $\;$ $\left(x + 7\right) \left(x + 9\right) + \left(y - 1\right) \left(y + 3\right) = 11$

i.e. $\;$ $x^2 + 16x + 63 + y^2 + 2y - 3 = 11$

i.e. $\;$ $x^2 + 16x + y^2 + 2y + 49 = 0$

Completing the square, we have,

$\left(x^2 + 16x + 64\right) + \left(y^2 + 2y + 1\right) + 49 - 64 - 1 = 0$

i.e. $\;$ $\left(x + 8\right)^2 + \left(y + 1\right)^2 - 16 = 0$

i.e. $\;$ $\left(x + 8\right)^2 + \left(y + 1\right)^2 = \left(4\right)^2$

Hence, the given equation represents a real circle whose center is $\left(-8, -1\right)$ and

radius is $= \sqrt{16} = 4$

Coordinate Geometry - Circle

Determine the nature of the loci $\;$ $x^2 + y^2 + 4x + 6y + 16 = 0$

Equation of given locus is: $\;$ $x^2 + y^2 + 4x + 6y + 16 = 0$

Completing the square, we have,

$\left(x^2 + 4x + 4\right) + \left(y^2 + 6y + 9\right) - 4 - 9 + 16 = 0$

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

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

Hence, the given equation represents an imaginary circle whose center is $\left(-2, -3\right)$ and

radius is $= \sqrt{-3} = i \sqrt{3}$ $\;$ where $\;$ $i = \sqrt{-1}$

Coordinate Geometry - Circle

Find the center and radius of the circle whose equation is $\;$ $\left(\ell x + my + 3 \ell\right)^2 + \left(mx - \ell y - 3m\right)^2 = 24 \left(\ell^2 + m^2\right)$

Given equation of circle: $\;$ $\left(\ell x + my + 3 \ell\right)^2 + \left(mx - \ell y - 3m\right)^2 = 24 \left(\ell^2 + m^2\right)$

i.e. $\;$ $\ell^2 x^2 + m^2 y^2 + 9 \ell^2 + 2 \ell m x y + 6 \ell m y + 6 \ell^2 x$

$\hspace{1cm}$ $+ m^2 x^2 + \ell^2 y^2 + 9 m^2 - 2 \ell m x y + 6 \ell m y - 6 m^2 x = 24 \ell^2 + 24 m^2$

i.e. $\;$ $\left(\ell^2 + m^2\right)x^2 + \left(\ell^2 + m^2\right)y^2 + \left(6 \ell^2 - 6 m^2\right)x + 12 \ell m y - \left(24 \ell^2 + 24 m^2\right) = 0$

i.e. $\;$ $x^2 + y^2 + \left[\dfrac{6\left(\ell^2 - m^2\right)}{\ell^2 + m^2}\right]x + \left[\dfrac{12 \ell m}{\ell^2 + m^2}\right]y - 24 = 0$

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

$g = \dfrac{3 \left(\ell^2 - m^2\right)}{\ell^2 + m^2}$, $\;$ $f = \dfrac{6 \ell m}{\ell^2 + m^2}$, $\;$ $c = -24$

center of the circle $= \left(-g, -f\right) = \left(\dfrac{-3 \left(\ell^2 - m^2\right)}{\ell^2 + m^2}, \dfrac{-6 \ell m}{\ell^2 + m^2}\right)$

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

i.e. $\;$ $r = \sqrt{\left(\dfrac{3 \left(\ell^2 - m^2\right)}{\ell^2 + m^2}\right)^2 + \left(\dfrac{6 \ell m}{\ell^2 + m^2}\right)^2 + 24}$

i.e. $\;$ $r = \dfrac{\sqrt{9 \ell^4 + 9 m^4 - 18 \ell^2 m^2 + 36 \ell^2 m^2 + 24 \ell^4 + 24 m^4 + 48 \ell^2 m^2}}{\ell^2 + m^2}$

i.e. $\;$ $r = \dfrac{\sqrt{33 \ell^4 + 33 m^4 + 66 \ell^2 m^2}}{\ell^2 + m^2}$

i.e. $\;$ $r = \dfrac{\sqrt{33 \left(\ell^4 + 2 \ell^2 m^2 + m^4\right)}}{\ell^2 + m^2}$

i.e. $\;$ $r = \dfrac{\sqrt{33 \left(\ell^2 + m^2\right)^2}}{\ell^2 + m^2}$

i.e. $\;$ $r = \sqrt{33}$

Coordinate Geometry - Circle

Find the center and radius of the circle whose equation is $\;$ $\left(x - 1\right) \left(x - 3\right) + \left(y - 2\right) \left(y - 4\right) = 0$

Given equation of circle: $\;$ $\left(x - 1\right) \left(x - 3\right) + \left(y - 2\right) \left(y - 4\right) = 0$

i.e. $\;$ $x^2 - 4x + 3 + y^2 - 6y + 8 = 0$

i.e. $\;$ $x^2 - 4x + y^2 - 6y + 11 = 0$

Completing the squares, we get

$\left(x^2 - 4x + 4\right) + \left(y^2 - 6y + 9\right) + 11 - 4 - 9 = 0$

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

Comparing with the standard equation of circle $\;$ $\left(x - h\right)^2 + \left(y - k\right)^2 = r^2$, $\;$ we have

$h = 2, \; \; k = 3, \;\; r^2 = 2$

center of the circle $= \left(h, k\right) = \left(2, 3\right)$

radius of the circle $= r = \sqrt{2}$

Coordinate Geometry - Circle

Find the center and radius of the circle whose equation is $\;$ $x^2 + y^2 + \dfrac{4}{7}x - \dfrac{6}{7}y = \dfrac{36}{49}$

Given equation of circle: $\;$ $x^2 = y^2 + \dfrac{4}{7}x - \dfrac{6}{7}y = \dfrac{36}{49}$

i.e. $\;$ $x^2 + y^2 + \dfrac{4}{7}x - \dfrac{6}{7}y - \dfrac{36}{49} = 0$

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

$2g = \dfrac{4}{7}$ $\implies$ $g = \dfrac{2}{7}$, $\;$ $2f = \dfrac{-6}{7}$ $\implies$ $f = \dfrac{-3}{7}$, $\;$ $c = \dfrac{-36}{49}$

Center of circle $= \left(-g, -f\right) = \left(\dfrac{-2}{7}, \dfrac{3}{7}\right)$

Radius of circle $= r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(\dfrac{2}{7}\right)^2 + \left(\dfrac{-3}{7}\right)^2 + \dfrac{36}{49}}$

i.e. $\;$ $r = \sqrt{\dfrac{4}{49} + \dfrac{9}{49} + \dfrac{36}{49}} = \sqrt{\dfrac{49}{49}} = 1$

Coordinate Geometry - Circle

Find the equation of the circle whose center is $\left(-p, -q\right)$ and radius is $\sqrt{p^2 + q^2}$.

Center $= \left(h, k\right) = \left(-p, -q\right)$

Radius $= r = \sqrt{p^2 + q^2}$

Equation of circle is $\;$ $\left(x - h\right)^2 + \left(y - k\right)^2 = r^2$

i.e. $\;$ $\left(x + p\right)^2 + \left(y + q\right)^2 = \left(\sqrt{p^2 + q^2}\right)^2$

i.e. $\;$ $x^2 + 2px + p^2 + y^2 + 2qy + q^2 = p^2 + q^2$

i.e. $\;$ $x^2 + y^2 + 2px + 2qy = 0$

Coordinate Geometry - Circle

Find the equation of the circle whose center is $\left(-7, -3\right)$ and radius is $10$.

Center $= \left(h, k\right) = \left(-7, -3\right)$

Radius $= r = 10$

Equation of circle is $\;$ $\left(x - h\right)^2 + \left(y - k\right)^2 = r^2$

i.e. $\;$ $\left(x + 7\right)^2 + \left(y + 3\right)^2 = 10^2$

i.e. $\;$ $x^2 + 14x + 49 + y^2 + 6y + 9 = 100$

i.e. $\;$ $x^2 + y^2 + 14x + 6y - 42 = 0$

Coordinate Geometry - Straight Line

What does the equation $\;$ $4x - 7y + 11 = 0$ $\;$ become when the origin is shifted to $\left(-3, -4\right)$.

Given equation of line: $\;$ $4x - 7y + 11 = 0$ $\;\;\; \cdots \; (1)$

The required equation is obtained by replacing $x$ by $\left(x - 3\right)$ and $y$ by $\left(y - 4\right)$ in equation $(1)$.

$\therefore \;$ The required equation is

$4 \left(x - 3\right) - 7 \left(y - 4\right) + 11 = 0$

i.e. $\;$ $4x - 12 - 7y + 28 + 11 = 0$

i.e. $\;$ $4x - 7y + 27 = 0$

Coordinate Geometry - Straight Line

Find the angle between the straight lines joining the origin to the intersections of the line $\;$ $y = 2x + 1$, $\;$ with the curve $\;$ $x^2 - 3xy + 2y^2 -4x -3y+ 1 = 0$

Equation of given curve is: $\;$ $x^2 - 3xy + 2y^2 -4x -3y+ 1 = 0$ $\;\;\; \cdots \; (1)$

Equation of given curve is: $\;$ $y = 2x + 1$ $\;\;\; \cdots \; (2)$

Equation $(2)$ can be written as

$y - 2x = 1$ $\;\;\; \cdots \; (2a)$

$\therefore \;$ Making equation $(1)$ homogeneous in $x$ and $y$ using equation $(2a)$ we get,

$x^2 - 3xy + 2y^2 - 4 \left(y - 2x\right)x - 3 \left(y - 2x\right)y + 1 \times \left(y - 2x\right)^2 = 0$

i.e. $\;$ $x^2 -3xy + 2y^2 - 4xy + 8x^2 - 3y^2 + 6xy + y^2 + 4x^2 - 4xy = 0$

i.e. $\;$ $13x^2 - 5xy + 0y^2 = 0$ $\;\;\; \cdots \; (3)$

Since equation $(3)$ is homogeneous and of the second degree in $x$ and $y$, it represents a pair of straight lines passing through the origin.

Comparing equation $(3)$ with the standard equation $\;$ $ax^2 + 2hxy + by^2 = 0$ $\;$ gives

$a = 13, \; h = \dfrac{-5}{2}, \; b = 0$

Let $\theta$ be the angle between the lines given by equation $(3)$.

Then, $\;$ $\tan \theta = \dfrac{2 \sqrt{h^2 - ab}}{a + b}$

Here, since $b = 0$, $\;$ $\tan \theta = \dfrac{2h}{a}$

i.e. $\;$ $\tan \theta = \dfrac{2 \times \left(\dfrac{-5}{2}\right)}{13} = \dfrac{-5}{13}$

$\implies$ $\theta = \tan^{-1} \left(\dfrac{-5}{13}\right)$

Coordinate Geometry - Straight Line

Show that, if the straight lines joining the origin to the points of intersection of $\;$ $3x^2 - xy + 3y^2 + 2x - 3y + 4 = 0$ $\;$ and $\;$ $2x + 3y = k$ $\;$ are at right angles, then $\;$ $6k^2 - 5k + 52 = 0$

Equation of given lines is: $\;$ $3x^2 - xy + 3y^2 + 2x - 3y + 4 = 0$ $\;\;\; \cdots \; (1)$

and $\;$ $2x + 3y = k$ $\;\;\; \cdots \; (2)$

Equation $(2)$ can be written as

$\dfrac{2x + 3y}{k} = 1$ $\;\;\; \cdots \; (2a)$

$\therefore \;$ Making equation $(1)$ homogeneous in $x$ and $y$ using equation $(2a)$ we get,

$3x^2 - xy + 3y^2 + 2 \left(\dfrac{2x + 3y}{k}\right)x - 3 \left(\dfrac{2x + 3y}{k}\right)y + 4 \times \left(\dfrac{2x + 3y}{k}\right)^2 = 0$

i.e. $\;$ $3k^2 x^2 - k^2 xy + 3k^2 y^2 + 4kx^2 + 6kxy -6kxy - 9ky^2 + 16x^2 + 36y^2 + 48xy = 0$

i.e. $\;$ $\left(3k^2 + 4k + 16\right)x^2 + \left(-k^2 + 48\right)xy + \left(3k^2 - 9k + 36\right)y^2 = 0$ $\;\;\; \cdots \; (3)$

Since equation $(3)$ is homogeneous and of the second degree in $x$ and $y$, it represents a pair of straight lines passing through the origin.

Lines represented by equation $(3)$ will be at right angles if

$\text{coefficient of } x^2 + \text{ coefficient of } y^2 = 0$

i.e. $\;$ $3k^2 + 4k + 16 + 3k^2 - 9k + 36 = 0$

i.e. $\;$ $6k^2 - 5k + 52 = 0$

Coordinate Geometry - Straight Line

Find the value of $k$ so that the pair of straight lines joining the origin to the points of intersection of $\;$ $x^2 + y^2 + 4x - 6y - 36 = 0$ $\;$ and the line $\;$ $y - x - k = 0$ $\;$ may be at right angles.

Equation of given lines is: $\;$ $x^2 + y^2 + 4x - 6y - 36 = 0$ $\;\;\; \cdots \; (1)$

and $\;$ $y - x - k = 0$ $\;\;\; \cdots \; (2)$

The combined equation to the pair of lines joining the origin to the points of intersection of equations $(1)$ and $(2)$ can be obtained by making equation $(1)$ homogeneous in $x$ and $y$ by using equation $(2)$.

Equation $(2)$ can be written as

$y - x = k$

i.e. $\;$ $\dfrac{y - x}{k} = 1$ $\;\;\; \cdots \; (2a)$

$\therefore \;$ Making equation $(1)$ homogeneous in $x$ and $y$ using equation $(2a)$ we get,

$x^2 + y^2 + 4 \left(\dfrac{y - x}{k}\right)x - 6 \left(\dfrac{y - x}{k}\right)y - 36 \times \left(\dfrac{y - x}{k}\right)^2 = 0$

i.e. $\;$ $k^2 x^2 + k^2 y^2 + 4kxy - 4kx^2 -6ky^2 + 6kxy - 36y^2 - 36x^2 + 72xy = 0$

i.e. $\;$ $\left(k^2 -4k - 36\right)x^2 + \left(10k + 72\right)xy + \left(k^2 -6k - 36\right)y^2 = 0$ $\;\;\; \cdots \; (3)$

Equation $(3)$ is a combination of equations $(1)$ and $(2a)$.

$\therefore \;$ Equation $(3)$ will be satisfied by the coordinates of all the points by which equations $(1)$ and $(2a)$ are satisfied simultaneously.

Since equation $(3)$ is homogeneous and of the second degree in $x$ and $y$, it represents a pair of straight lines passing through the origin.

Lines represented by equation $(3)$ will be at right angles if

$\text{coefficient of } x^2 + \text{ coefficient of } y^2 = 0$

i.e. $\;$ $k^2 - 4k - 36 + k^2 - 6k - 36 = 0$

i.e. $\;$ $2k^2 - 10k - 72 = 0$

i.e. $\;$ $k^2 - 5k - 36 = 0$

$\implies$ $k = 9$ $\;$ or $\;$ $k = -4$

Coordinate Geometry - Straight Line

Prove that the equation of the pair of lines through the origin perpendicular to the pair whose equation is $\;$ $ax^2 + 2hxy + by^2 = 0$ $\;$ is $\;$ $bx^2 - 2hxy + ay^2 = 0$.

Equation of given pair of lines is: $\;$ $ax^2 + 2hxy + by^2 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $m_1$ $\;$ and $\;$ $m_2$ $\;$ be the slopes of the two lines given by equation $(1)$.

Then, $\;$ $m_1 + m_2 = \dfrac{-2h}{b}$ $\;\;\; \cdots \; (2a)$

$m_1 m_2 = \dfrac{a}{b}$ $\;\;\; \cdots \; (2b)$

and $\;$ $m_1 - m_2 = \dfrac{2 \sqrt{h^2 - ab}}{\left|b\right|}$ $\;\;\; \cdots \; (2c)$

From equations $(2a)$ and $(2c)$,

$2m_1 = \dfrac{-2h}{b} + \dfrac{2 \sqrt{h^2 - ab}}{b}$

$\implies$ $m_1 = \dfrac{-h + \sqrt{h^2 - ab}}{b}$ $\;\;\; \cdots \; (3a)$

and $\;$ $2m_2 = \dfrac{-2h}{b} - \dfrac{2 \sqrt{h^2 - ab}}{b}$

$\implies$ $m_2 = \dfrac{-h - \sqrt{h^2 - ab}}{b}$ $\;\;\; \cdots \; (3b)$

Let $\;$ $m_1'$ $\;$ $m_2'$ $\;$ be the slopes of the lines perpendicular to those given by equation $(1)$.

Then we have,

$m_1' = \dfrac{-1}{m_1} = \dfrac{-b}{-h + \sqrt{h^2 - ab}}$ $\;\;\; \cdots \; (4a)$ $\;\;$ [in view of equation $(3a)$]

and $\;$ $m_2' = \dfrac{-1}{m_2} = \dfrac{-b}{-h - \sqrt{h^2 - ab}} = \dfrac{b}{h + \sqrt{h^2 - ab}}$ $\;\;\; \cdots \; (4b)$ $\;\;$ [in view of equation $(3b)$]

We have from equations $(4a)$ and $(4b)$,

$m_1' + m_2' = \dfrac{-b}{-h + \sqrt{h^2 - ab}} + \dfrac{b}{h + \sqrt{h^2 - ab}}$

i.e. $\;$ $m_1' + m_2' = -b \left[\dfrac{h + \sqrt{h^2 - ab} - \left(-h + \sqrt{h^2 - ab}\right)}{\left(-h + \sqrt{h^2 - ab}\right) \left(h + \sqrt{h^2 - ab}\right)}\right]$

i.e. $\;$ $m_1' + m_2' = - b \left[\dfrac{2h}{h^2 - ab - h^2}\right] = \dfrac{2bh}{ab}$

i.e. $\;$ $m_1' + m_2' = \dfrac{2h}{a}$ $\;\;\; \cdots \; (5a)$

and $\;$ $m_1' \times m_2' = \left(\dfrac{-b}{-h + \sqrt{h^2 - ab}}\right) \left(\dfrac{b}{h + \sqrt{h^2 - ab}}\right)$

i.e. $\;$ $m_1' m_2' = \dfrac{-b^2}{h^2 - ab - h^2} = \dfrac{-b^2}{-ab}$

i.e. $\;$ $m_1' m_2' = \dfrac{b}{a}$ $\;\;\; \cdots \; (5b)$

In view of equations $(5a)$ and $(5b)$, the combined equation of pair of lines with slopes $m_1'$ and $m_2'$ is

$bx^2 - 2hxy + ay^2 = 0$ $\;\;\; \cdots \; (6)$

i.e. $\;$ The equation of pair of lines through the origin perpendicular to the pair given by equation $(1)$ are given by equation $(6)$.

Coordinate Geometry - Straight Line

Prove that the equation $\;$ $y^3 - x^3 + 3xy \left(y - x\right) = 0$ $\;$ represents three straight lines equally inclined to one another.

Given equation is: $\;$ $y^3 - x^3 + 3xy \left(y - x\right) = 0$ $\;\;\; \cdots \; (1a)$

Equation $(1a)$ can be written as

$\left(y - x\right) \left(y^2 + xy + x^2\right) + 3xy \left(y - x\right) = 0$

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

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

Equation $(1b)$ represents the pair of lines

$y - x = 0$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $x^2 + 4xy + y^2 = 0$ $\;\;\; \cdots \; (2b)$

Equation $(2a)$ can be written as $\;\;$ $y = x$

Comparing with the standard equation $\;$ $y = mx + c$ $\;$ (c is the intercept on the Y axis) $\;$ gives

slope of equation $(2a)$ $= m = 1$ $\;\;\; \cdots \; (3a)$

Comparing equation $(2b)$ with the standard equation $\;\;$ $ax^2 + 2hxy + by^2 = 0$ $\;$ gives

$a = 1, \; h = 2, \; b = 1$ $\;\;\; \cdots \; (3b)$

Let $\theta$ be the angle between the pair of straight lines given by equation $(2b)$.

Then,

$\begin{aligned} \tan \theta & = \dfrac{2 \sqrt{h^2 - ab}}{a + b} \\\\ & = \dfrac{2 \sqrt{2^2 - 1 \times 1}}{1 + 1} \\\\ & = \dfrac{2 \sqrt{3}}{2} = \sqrt{3} \end{aligned}$

$\implies$ $\theta = \tan^{-1} \left(\sqrt{3}\right) = 60^\circ$ $\;\;\; \cdots \; (4a)$

Let $m_1$ and $m_2$ be the slopes of the lines given by equation $(2b)$.

Then, $\;$ $m_1 + m_2 = \dfrac{-2h}{b} = \dfrac{-2 \times 2}{1} = -4$ $\;\;\; \cdots \; (5a)$

and $\;$ $m_1 - m_2 = \dfrac{2 \sqrt{h^2 - ab}}{\left|b\right|} = \dfrac{2 \sqrt{2^2 - 1 \times 1}}{\left|1\right|} = 2 \sqrt{3}$ $\;\;\; \cdots \; (5b)$

From equations $(5a)$ and $(5b)$,

$2 m_1 = -4 + 2 \sqrt{3}$ $\implies$ $m_1 = \sqrt{3} - 2$ $\;\;\; \cdots \; (6a)$

and $\;$ $m_2 = -4 - m_1 = -4 - \left(\sqrt{3} - 2\right) = -\sqrt{3} - 2$ $\;\;\; \cdots \; (6b)$

Let $\phi$ be the angle between the lines with slopes $m$ and $m_1$.

Then,

$\begin{aligned} \tan \phi & = \left|\dfrac{m - m_1}{1 + m m_1}\right| \\\\ & = \left|\dfrac{1 - \left(\sqrt{3} - 2\right)}{1 + 1 \times \left(\sqrt{3} - 2\right)}\right| \;\; \left[\text{by equations (3a) and (6a)}\right] \\\\ & = \left|\dfrac{3 - \sqrt{3}}{1 - 2 + \sqrt{3}}\right| \\\\ & = \dfrac{3 - \sqrt{3}}{\sqrt{3} - 1} \\\\ & = \dfrac{\left(3 - \sqrt{3}\right) \left(\sqrt{3} + 1\right)}{\left(\sqrt{3} - 1\right) \left(\sqrt{3} + 1\right)} \\\\ & = \dfrac{3 \sqrt{3} + 3 - 3 - \sqrt{3}}{3 - 1} = \dfrac{2 \sqrt{3}}{2} = \sqrt{3} \end{aligned}$

$\implies$ $\phi = \tan^{-1} \left(\sqrt{3}\right) = 60^\circ$ $\;\;\; \cdots \; (4b)$

Let $\gamma$ be the angle between the lines with slopes $m$ and $m_2$.

Then,

$\begin{aligned} \tan \gamma & = \left|\dfrac{m - m_2}{1 + m m_2}\right| \\\\ & = \left|\dfrac{1 - \left(-\sqrt{3} - 2\right)}{1 + 1 \times \left(-\sqrt{3} - 2\right)}\right| \;\; \left[\text{by equations (3a) and (6b)}\right] \\\\ & = \left|\dfrac{3 + \sqrt{3}}{1 - 2 - \sqrt{3}}\right| \\\\ & = \dfrac{3 + \sqrt{3}}{\sqrt{3} + 1} \\\\ & = \dfrac{\left(3 + \sqrt{3}\right) \left(\sqrt{3} - 1\right)}{\left(\sqrt{3} - 1\right) \left(\sqrt{3} + 1\right)} \\\\ & = \dfrac{3 \sqrt{3} - 3 + 3 - \sqrt{3}}{3 - 1} = \dfrac{2 \sqrt{3}}{2} = \sqrt{3} \end{aligned}$

$\implies$ $\gamma = \tan^{-1} \left(\sqrt{3}\right) = 60^\circ$ $\;\;\; \cdots \; (4c)$

$\therefore \;$ We have from equations $(4a)$, $(4b)$ and $(4c)$, $\;$ $\theta = \phi = \gamma$

$\implies$ The three straight lines given by equation $(1a)$ are equally inclined to one another.

Hence proved.

Coordinate Geometry - Straight Line

Find the value of $k$ in order that the equation $\;$ $2x^2 + xy - y^2 + kx + 6y - 9 = 0$ $\;$ may represent a pair of straight lines.

Given equation is: $\;$ $2x^2 + xy - y^2 + kx + 6y - 9 = 0$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the general second degree equation

$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives

$a = 2, \; h = \dfrac{1}{2}, \; b = -1, \; g = \dfrac{k}{2}, \; f = 3, \; c = -9$

Equation $(2)$ represents a pair of straight lines if

$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ $\;\;\; \cdots \; (3)$

Substituting the values of $\;$ $a, \; b, \; c, \; f, \; g, \; h \;$ in equation $(3)$ gives

$2 \times \left(-1\right) \times \left(-9\right) + 2 \times 3 \times \dfrac{k}{2} \times \dfrac{1}{2} - 2 \times 3^2 - \left(-1\right) \times \left(\dfrac{k}{2}\right)^2 - \left(-9\right) \times \left(\dfrac{1}{2}\right)^2 = 0$

i.e. $\;$ $18 + \dfrac{3k}{2} - 18 + \dfrac{k^2}{4} + \dfrac{9}{4} = 0$

i.e. $\;$ $k^2 + 6k + 9 = 0$

i.e. $\;$ $\left(k + 3\right)^2 = 0$

$\implies$ $k = -3$

Coordinate Geometry - Straight Line

Find the value of $k$ in order that the equation $\;$ $12x^2 + kxy + 2y^2 + 11x - 5y + 2 = 0$ $\;$ may represent a pair of straight lines.

Given equation is: $\;$ $12x^2 + kxy + 2y^2 + 11x - 5y + 2 = 0$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the general second degree equation

$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives

$a = 12, \; h = \dfrac{k}{2}, \; b = 2, \; g = \dfrac{11}{2}, \; f = \dfrac{-5}{2}, \; c = 2$

Equation $(2)$ represents a pair of straight lines if

$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ $\;\;\; \cdots \; (3)$

Substituting the values of $\;$ $a, \; b, \; c, \; f, \; g, \; h \;$ in equation $(3)$ gives

$12 \times 2 \times 2 + 2 \times \left(\dfrac{-5}{2}\right) \times \dfrac{11}{2} \times \dfrac{k}{2} - 12 \times \left(\dfrac{-5}{2}\right)^2 - 2 \times \left(\dfrac{11}{2}\right)^2 - 2 \times \left(\dfrac{k}{2}\right)^2 = 0$

i.e. $\;$ $48 - \dfrac{55 k}{4} - 75 - \dfrac{121}{2} - \dfrac{k^2}{2} = 0$

i.e. $\;$ $\dfrac{-175}{2} - \dfrac{55k}{4} - \dfrac{k^2}{2} = 0$

i.e. $\;$ $2k^2 + 55k + 350 = 0$

$\implies$ $k = -10$ $\;$ or $\;$ $k = -17.5$

Coordinate Geometry - Straight Line

Find the value of $k$ in order that the equation $\;$ $10x^2 + xy + ky^2 + x + 5y - 3 = 0$ $\;$ may represent a pair of straight lines.

Given equation is: $\;$ $10x^2 + xy + ky^2 + x + 5y - 3 = 0$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the general second degree equation

$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives

$a = 10, \; h = \dfrac{1}{2}, \; b = k, \; g = \dfrac{1}{2}, \; f = \dfrac{5}{2}, \; c = -3$

Equation $(2)$ represents a pair of straight lines if

$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ $\;\;\; \cdots \; (3)$

Substituting the values of $\;$ $a, \; b, \; c, \; f, \; g, \; h \;$ in equation $(3)$ gives

$10 \times k \times \left(-3\right) + 2 \times \dfrac{5}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} - 10 \times \left(\dfrac{5}{2}\right)^2 - k \times \left(\dfrac{1}{2}\right)^2 - \left(-3\right) \times \left(\dfrac{1}{2}\right)^2 = 0$

i.e. $\;$ $-30k - \dfrac{k}{4} + \dfrac{5}{4} - \dfrac{250}{4} + \dfrac{3}{4} = 0$

i.e. $\;$ $121 k = - 242$ $\implies$ $k = -2$

Coordinate Geometry - Straight Line

Show that the equation $\;$ $x^2 + 2xy + y^2 - 8ax - 8ay - 9a^2 = 0$ $\;$ represents a pair of parallel straight lines and find the perpendicular distance between them.

Given equation is: $\;$ $x^2 + 2xy + y^2 - 8ax - 8ay - 9a^2 = 0$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the general second degree equation

$Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives

$A = 1, \; H = 1, \; B = 1, \; G = -4a, \; F = -4a, \; C = -9a^2$

Equation $(2)$ represents a pair of parallel straight lines if

$H^2 = AB$ $\;$ and $\;$ $BG^2 = AF^2$

Now, $\;$ $H^2 = 1^2 = 1$, $\;\;$ $A \times B = 1 \times 1 = 1$

i.e. $\;$ $H^2 = AB$

and $\;$ $BG^2 = 1 \times \left(-4a\right)^2 = 16a^2$, $\;\;$ $AF^2 = 1 \times \left(-4a\right)^2 = 16a^2$

i.e. $\;$ $BG^2 = AF^2$

$\implies$ Equation $(1)$ represents a pair of parallel straight lines.

Consider the terms $\;\;$ $x^2 + 2xy + y^2$ $\;\;$ from equation $(1)$.

Now, $\;\;$ $x^2 + 2xy + y^2 = \left(x + y\right) \left(x + y\right)$

$\therefore \;$ Equation $(1)$ can be written as

$\left(x + y + \ell\right) \left(x + y + m\right) = 0$ $\;\;\; \cdots \; (2) \;$ where $\ell$ and $m$ are constants

Since equations $(1)$ and $(2)$ represent the same pair of straight lines,

comparing the coefficients of the $x$ terms gives

$\ell + m = -8a$ $\;\;\; \cdots \; (3a)$

comparing the constant terms gives

$\ell \times m = -9a^2$ $\;\;\; \cdots \; (3b)$

Now,

$\begin{aligned} \left(\ell - m\right)^2 & = \left(\ell + m\right)^2 - 4 \ell m \\\\ & = \left(-8a\right)^2 - 4 \times \left(-9a^2\right) \\\\ & = 64a^2 - 36a^2 = 28a^2 \end{aligned}$

$\therefore \;$ $\left|\ell - m\right| = 2a \sqrt{7}$ $\;\;\; \cdots \; (4)$

Equation $(1)$ is represented by the two parallel lines

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

Distance between the parallel lines given by equations $(5a)$ and $(5b)$ is

$= \dfrac{\left|\ell - m\right|}{\sqrt{\left(\text{coefficient of x}\right)^2 + \left(\text{coefficient of y}\right)^2}}$

$= \dfrac{2a \sqrt{7}}{\sqrt{1^2 + 1^2}}$

$= \dfrac{2a \sqrt{7}}{\sqrt{2}} = a \sqrt{14}$

Coordinate Geometry - Straight Line

Show that the equation $\;$ $9x^2 - 12xy + 4y^2 + 15x - 10y + 4 = 0$ $\;$ represents a pair of parallel straight lines.

Given equation is: $\;$ $9x^2 - 12xy + 4y^2 + 15x - 10y + 4 = 0$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the general second degree equation

$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives

$a = 9, \; h = -6, \; b = 4, \; g = \dfrac{15}{2}, \; f = -5, \; c = 4$

Equation $(2)$ represents a pair of parallel straight lines if

$h^2 = ab$ $\;$ and $\;$ $bg^2 = af^2$

Now, $\;$ $h^2 = \left(-6\right)^2 = 36$, $\;\;$ $a \times b = 9 \times 4 = 36$

i.e. $\;$ $h^2 = ab$

and $\;$ $bg^2 = 4 \times \left(\dfrac{15}{2}\right)^2 = 225$, $\;\;$ $af^2 = 9 \times \left(-5\right)^2 = 225$

i.e. $\;$ $bg^2 = af^2$

$\implies$ Equation $(1)$ represents a pair of parallel straight lines.

Coordinate Geometry - Straight Line

Prove that the equation $\;$ $x^2 - 5xy + 4y^2 + x + 2y - 2 = 0$ $\;$ represents two straight lines.

Given equation is: $\;$ $x^2 - 5xy + 4y^2 + x + 2y - 2 = 0$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the general second degree equation

$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives

$a = 1, \; h = \dfrac{-5}{2}, \; b = 4, \; g = \dfrac{1}{2}, \; f = 1, \; c = -2$

Equation $(2)$ represents a pair of straight lines if

$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ $\;$ i.e. $\;$ $\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0$

Substituting the values of $\;$ $a, \; b, \; c, \; f, \; g, \; h \;$ in the expression $\;$ $abc + 2fgh - af^2 - bg^2 - ch^2$ $\;$ we have,

$1 \times 4 \times \left(-2\right) + 2 \times 1 \times \dfrac{1}{2} \times \left(\dfrac{-5}{2}\right) - 1 \times 1^2 - 4 \times \left(\dfrac{1}{2}\right)^2 - \left(-2\right) \times \left(\dfrac{-5}{2}\right)^2$

$= -8 - \dfrac{5}{2} -1 -1 + \dfrac{25}{2}$

$= -10 + 10 = 0$

i.e. $\;$ $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$

$\implies$ Equation $(1)$ represents a pair of straight lines.

Coordinate Geometry - Straight Line

Prove that the equation $\;$ $x^2 - 3xy + 2y^2 + 4x - 5y + 3 = 0$ $\;$ represents two straight lines.

Given equation is: $\;$ $x^2 - 3xy + 2y^2 + 4x - 5y + 3 = 0$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the general second degree equation

$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives

$a = 1, \; h = \dfrac{-3}{2}, \; b = 2, \; g = 2, \; f = \dfrac{-5}{2}, \; c = 3$

Equation $(2)$ represents a pair of straight lines if

$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$

Substituting the values of $\;$ $a, \; b, \; c, \; f, \; g, \; h \;$ in the expression $\;$ $abc + 2fgh - af^2 - bg^2 - ch^2$ $\;$ we have,

$1 \times 2 \times 3 + 2 \times \left(\dfrac{-5}{2}\right) \times 2 \times \left(\dfrac{-3}{2}\right) - 1 \times \left(\dfrac{-5}{2}\right)^2 - 2 \times 2^2 - 3 \times \left(\dfrac{-3}{2}\right)^2$

$= 6 + 15 - \dfrac{25}{4} - 8 - \dfrac{27}{4}$

$= 13 - \dfrac{52}{4} = 0$

i.e. $\;$ $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$

$\implies$ Equation $(1)$ represents a pair of straight lines.

Coordinate Geometry - Straight Line

Show that the two straight lines $\;$ $x^2 \left(\tan^2 \theta + \cos^2 \theta\right) - 2 xy \tan \theta + y^2 \sin^2 \theta = 0$ $\;$ make with the $X$ axis angles such that the difference of their tangents is $2$.

Equation of given pair of lines is:

$x^2 \left(\tan^2 \theta + \cos^2 \theta\right) - 2 xy \tan \theta + y^2 \sin^2 \theta = 0$ $\;\;\; \cdots \; (1a)$

Equation $(1a)$ can be written as

$x^2 \left(\dfrac{\tan^2 \theta + \cos^2 \theta}{\sin^2 \theta}\right) - 2 \dfrac{\tan \theta}{\sin^2 \theta} xy + y^2 = 0$ $\;\;\; \cdots \; (1b)$

Let the lines represented by equation $(1b)$ make angles $\phi_1$ and $\phi_2$ with the $X$ axis.

Then the combined equation of the lines given by equation $(1b)$ can be written as

$\left(y - x \tan \phi_1\right) \left(y - x \tan \phi_2\right) = 0$

i.e. $\;$ $y^2 - xy \tan \phi_2 - xy \tan \phi_1 + x^2 \tan \phi_1 \tan \phi_2 = 0$

i.e. $\;$ $y^2 - xy \left(\tan \phi_1 + \tan \phi_2\right) + x^2 \tan \phi_1 \tan \phi_2 = 0$ $\;\;\; \cdots \; (2)$

Since equations $(1b)$ and $(2)$ are identical equations, we have,

$\tan \phi_1 \tan \phi_2 = \dfrac{\tan^2 \theta + \cos^2 \theta}{\sin^2 \theta}$ $\;\;\; \cdots \; (3a)$ $\;$ and

$\tan \phi_1 + \tan \phi_2 = \dfrac{2 \tan \theta}{\sin^2 \theta}$ $\;\;\;\; \cdots \; (3b)$

Difference of tangents of the angles made by the lines given by equation $(1a)$ is $= \tan \phi_1 - \tan \phi_2$

Now,

$\begin{aligned} \left(\tan \phi_1 - \tan \phi_2\right)^2 & = \left(\tan \phi_1 + \tan \phi_2\right)^2 - 4 \tan \phi_1 \tan \phi_2 \\\\ & = \left(\dfrac{2 \tan \theta}{\sin^2 \theta}\right)^2 - 4 \times \left(\dfrac{\tan^2 \theta + \cos^2 \theta}{\sin^2 \theta}\right) \\\\ & = \dfrac{4 \tan^2 \theta}{\sin^4 \theta} - 4 \times \left(\dfrac{\tan^2 \theta \sin^2 \theta + \sin^2 \theta \cos^2 \theta}{\sin^4 \theta}\right) \\\\ & = \dfrac{4 \tan^2 \theta \left(1 - \sin^2 \theta\right) - 4 \sin^2 \theta \cos^2 \theta}{\sin^4 \theta} \\\\ & = \dfrac{4 \tan^2 \theta \cos^2 \theta - 4 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta} \\\\ & = \dfrac{4 \cos^2 \theta \left(\tan^2 \theta - \sin^2 \theta\right)}{\sin^4 \theta} \\\\ & = \dfrac{4 \cos^2 \theta \left(\dfrac{\sin^2 \theta - \sin^2 \theta \cos^2 \theta}{\cos^2 \theta}\right)}{\sin^4 \theta} \\\\ & = \dfrac{4 \sin^2 \theta \left(1 - \cos^2 \theta\right)}{\sin^4 \theta} \\\\ & = \dfrac{4 \times \sin^2 \theta \times \sin^2 \theta}{\sin^4 \theta} \end{aligned}$

i.e. $\;$ $\left(\tan \phi_1 - \tan \phi_2\right)^2 = 4$

$\implies$ $\tan \phi_1 - \tan \phi_2 = 2$

Coordinate Geometry - Straight Line

Find the equations of the bisectors of the angles between the pair of straight lines given by $\;$ $x^2 + 2xy \cot \alpha + y^2 = 0$

Equation of given pair of lines: $\;$ $x^2 + 2xy \cot \alpha + y^2 = 0$ $\;\;\; \cdots \; (1)$

Comparing with the standard equation of pair of lines: $\;$ $ax^2 + 2hxy + by^2 = 0$ $\cdots \; (2)$ gives

$a = 1, \;\; h = \cot \alpha, \;\; b = 1$

The combined equation of the bisectors of equation $(2)$ is $\;\;$ $\dfrac{x^2 - y^2}{a - b} = \dfrac{xy}{h}$ $\;$

i.e. $\;$ $x^2 - y^2 = \left(a - b\right) \dfrac{xy}{h}$

$\therefore \;$ The combined equation of the bisectors of equation $(1)$ is

$x^2 - y^2 = \left(1 - 1\right) \times \dfrac{xy}{\cot \alpha}$

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

Coordinate Geometry - Straight Line

Find the equations of the bisectors of the angles between the pair of straight lines given by $\;$ $6x^2 + 5xy - 6y^2 = 0$

Equation of given pair of lines: $\;$ $6x^2 + 5xy - 6y^2 = 0$ $\;\;\; \cdots \; (1)$

Comparing with the standard equation of pair of lines: $\;$ $ax^2 + 2hxy + by^2 = 0$ $\cdots \; (2)$ gives

$a = 6, \;\; h = \dfrac{5}{2}, \;\; b = -6$

The combined equation of the bisectors of equation $(2)$ is $\;\;$ $\dfrac{x^2 - y^2}{a - b} = \dfrac{xy}{h}$

$\therefore \;$ The combined equation of the bisectors of equation $(1)$ is

$\dfrac{x^2 - y^2}{6 - \left(-6\right)} = \dfrac{xy}{\dfrac{5}{2}}$

i.e. $\;$ $\dfrac{x^2 - y^2}{12} = \dfrac{2 xy}{5}$

i.e. $\;$ $5 \left(x^2 - y^2\right) - 24 xy = 0$

Coordinate Geometry - Straight Line

Find the acute angle between the pair of straight lines $\;$ $6y^2 - xy - x^2 + 30y + 36 = 0$

Equation of given pair of lines: $\;$ $6y^2 - xy - x^2 + 30y + 36 = 0$

Comparing with the standard equation of pair of lines: $\;$ $ax^2 + 2hxy + by^2 + 2 gx + 2 fy + c = 0$ $\;$ gives

$a = -1, \;\; h = \dfrac{-1}{2}, \;\; b = 6$

Let $\theta$ be the angle between the given pair of lines.

Then, $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{h^2 - ab}}{a + b}\right|$

i.e. $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{\left(\dfrac{-1}{2}\right)^2 - \left(-1\right) \times 6}}{-1 + 6}\right|$

i.e. $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{\dfrac{1}{4} + 6}}{5}\right| = \dfrac{5}{5} = 1$

$\implies$ $\theta = \tan^{-1} \left(1\right) = 45^\circ$

Coordinate Geometry - Straight Line

Find the acute angle between the pair of straight lines $\;$ $x^2 + 2xy \sec x + y^2 = 0$

Equation of given pair of lines: $\;$ $x^2 + 2xy \sec x + y^2 = 0$

Comparing with the standard equation of pair of lines: $\;$ $ax^2 + 2hxy + by^2 = 0$ $\;$ gives

$a = 1, \;\; h = \sec x, \;\; b = 1$

Let $\theta$ be the angle between the given pair of lines.

Then, $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{h^2 - ab}}{a + b}\right|$

i.e. $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{\sec^2 x - 1 \times 1}}{1 + 1}\right|$

i.e. $\;$ $\tan \theta = \left|\sqrt{\sec^2 x - 1}\right| = \sqrt{\tan^2 x} = \tan x$

$\implies$ $\theta = \tan^{-1} \left(\tan x\right) = x$

Coordinate Geometry - Straight Line

Find the acute angle between the pair of straight lines $\;$ $x^2 - 7xy + 12y^2 = 0$

Equation of given pair of lines: $\;$ $x^2 - 7xy + 12y^2 = 0$

Comparing with the standard equation of pair of lines: $\;$ $ax^2 + 2hxy + by^2 = 0$ $\;$ gives

$a = 1, \;\; h = \dfrac{-7}{2}, \;\; b = 12$

Let $\theta$ be the angle between the given pair of lines.

Then, $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{h^2 - ab}}{a + b}\right|$

i.e. $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{\left(\dfrac{-7}{2}\right)^2 - 1 \times 12}}{1 + 12}\right|$

i.e. $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{\dfrac{49}{4} - 12}}{13}\right| = \left|\dfrac{1}{13}\right| = \dfrac{1}{13}$

$\implies$ $\theta = \tan^{-1} \left(\dfrac{1}{13}\right)$

Coordinate Geometry - Straight Line

Find the acute angle between the pair of straight lines $\;$ $x^2 - xy - 6y^2 = 0$

Equation of given pair of lines: $\;$ $x^2 - xy - 6y^2 = 0$

Comparing with the standard equation of pair of lines: $\;$ $ax^2 + 2hxy + by^2 = 0$ $\;$ gives

$a = 1, \;\; h = \dfrac{-1}{2}, \;\; b = -6$

Let $\theta$ be the angle between the given pair of lines.

Then, $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{h^2 - ab}}{a + b}\right|$

i.e. $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{\left(\dfrac{-1}{2}\right)^2 - 1 \times \left(-6\right)}}{1 - 6}\right|$

i.e. $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{\dfrac{1}{4} + 6}}{-5}\right| = \left|\dfrac{5}{-5}\right| = \left|-1\right| = 1$

$\implies$ $\theta = \tan^{-1} \left(1\right) = 45^\circ$

Coordinate Geometry - Straight Line

Find the loci represented by the equation $\;$ $\left(x - a\right)^2 + \left(y - b\right)^2 = 0$

The given equation is $\;$ $\left(x - a\right)^2 + \left(y - b\right)^2 = 0$

Since the sum of squares of two real quantities cannot be zero unless each of them is zero, therefore, we must have,

$x - a = 0$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $y - b = 0$ $\;\;\; \cdots \; (2)$ $\;$ simultaneously.

$\because \;$ Both equations $(1)$ and $(2)$ are satisfied simultaneously,

the given equation represents the point of intersection of the loci given by equations $(1)$ and $(2)$.

$\therefore \;$ The given equation represents the point $\;$ $\left(a, b\right)$ $\;$ which is the point of intersection of lines given by equations $(1)$ and $(2)$.

Coordinate Geometry - Straight Line

Find the loci represented by the equation $\;$ $x^2 y + xy = 0$

The given equation is $\;$ $x^2 y + xy = 0$

The given equation can be written as $\;$ $xy \left(x + 1\right) = 0$

$\implies$ $xy = 0$ $\;$ or $\;$ $x + 1 = 0$

i.e. $\;$ $x = 0$ $\;$ or $\;$ $y = 0$ $\;$ or $\;$ $x = -1$

Hence, the given equation represents the lines

$x = 0$ $\;$ or $\;$ $y = 0$ $\;$ and $\;$ $x = -1$

Coordinate Geometry - Straight Line

Find the loci represented by the equation $\;$ $x^2 - y^2 = 0$

The given equation is $\;$ $x^2 - y^2 = 0$

The given equation can be written as $\;$ $\left(x + y\right) \left(x - y\right) = 0$

$\implies$ $x + y = 0$ $\;$ or $\;$ $x - y = 0$

i.e. $\;$ $x = -y$ $\;$ or $x = y$

Hence, the given equation represents the two lines

$x = -y$ $\;$ and $\;$ $x = y$

Coordinate Geometry - Straight Line

Find the equation of the line joining the points $\left(1, \dfrac{\pi}{4}\right)$ and $\left(2, \dfrac{\pi}{2}\right)$.

Let $\;$ $A \left(r_1, \theta_1\right) = \left(1, \dfrac{\pi}{4}\right)$, $\;$ $B \left(r_2, \theta_2\right) = \left(2, \dfrac{\pi}{2}\right)$

Let $P \left(r, \theta\right)$ be any point on the line $AB$.

Then the equation of line $AB$ is

$r_1 r \sin \left(\theta - \theta_1\right) + r r_2 \sin \left(\theta_2 - \theta\right) + r_2 r_1 \sin \left(\theta_1 - \theta_2\right) = 0$

i.e. $\;$ $1 \times r \sin \left(\theta - \dfrac{\pi}{4}\right) + r \times 2 \sin \left(\dfrac{\pi}{2} - \theta\right) + 1 \times 2 \sin \left(\dfrac{\pi}{4} - \dfrac{\pi}{2}\right) = 0$

i.e. $\;$ $r \left(\sin \theta \cos \dfrac{\pi}{4} - \cos \theta \sin \dfrac{\pi}{4}\right) + 2r \cos \theta + 2 \sin \left(\dfrac{- \pi}{4}\right) = 0$

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

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

Coordinate Geometry - Straight Line

Express the equation $\;$ $r^{\frac{1}{2}} = a^{\frac{1}{2}} \cos \left(\dfrac{\theta}{2}\right)$ $\;$ in Cartesian coordinates.

The relation between Cartesian coordinates $\left(x, y\right)$ and polar coordinates $\left(r, \theta\right)$ is

$x = r \cos \theta$, $\;\;$ $y = r \sin \theta$

$\implies$ $\cos \theta = \dfrac{x}{r}$, $\;\;$ $\sin \theta = \dfrac{y}{r}$

Now, $\;$ $\cos \theta = 2 \cos^2 \left(\dfrac{\theta}{2}\right) - 1$

$\therefore \;$ $\cos \left(\dfrac{\theta}{2}\right) = \sqrt{\dfrac{1 + \cos \theta}{2}}$

Given equation is $\;\;$ $r^{\frac{1}{2}} = a^{\frac{1}{2}} \cos \left(\dfrac{\theta}{2}\right)$

i.e. $\;$ $\sqrt{r} = \sqrt{a} \times \sqrt{\dfrac{1 + \cos \theta}{2}}$

i.e. $\;$ $\sqrt{r} = \sqrt{a} \times \sqrt{\dfrac{1 + \frac{x}{r}}{2}}$

i.e. $\;$ $\sqrt{r} = \sqrt{a} \times \sqrt{\dfrac{x + r}{2r}}$

i.e. $\;$ $r = \sqrt{a} \times \sqrt{\dfrac{x + r}{2}}$

i.e. $\;$ $2 r^2 = a \left(x + r\right)$ $\;\;\; \cdots \; (1)$

Now, $\;$ $r^2 = x^2 + y^2$ $\;\;\; \cdots \; (2a)$ $\implies$ $r = \sqrt{x^2 + y^2}$ $\;\;\; \cdots \; (2b)$

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

$2 \left(x^2 + y^2\right) = a \left(x + \sqrt{x^2 + y^2}\right)$

i.e. $\;$ $2 \left(x^2 + y^2\right) - ax = a \sqrt{x^2 + y^2}$

i.e. $\;$ $\left[2 \left(x^2 + y^2\right) - ax\right]^2 = a^2 \left(x^2 + y^2\right)$

$\therefore \;$ The given equation in Cartesian coordinates is $\;$ $\left[2 \left(x^2 + y^2\right) - ax\right]^2 = a^2 \left(x^2 + y^2\right)$

Coordinate Geometry - Straight Line

Express the equation $\;$ $r^2 \sin 2 \theta = a^2$ $\;$ in Cartesian coordinates.

The relation between Cartesian coordinates $\left(x, y\right)$ and polar coordinates $\left(r, \theta\right)$ is

$x = r \cos \theta$, $\;\;$ $y = r \sin \theta$

$\implies$ $\cos \theta = \dfrac{x}{r}$, $\;\;$ $\sin \theta = \dfrac{y}{r}$

Given equation is $\;\;$ $r^2 \sin 2\theta = a^2$

i.e. $\;$ $2 r^2 \sin \theta \cos \theta = a^2$

i.e. $\;$ $2 r^2 \times \dfrac{y}{r} \times \dfrac{x}{r} = a^2$

i.e. $\;$ $2 x y = a^2$

$\therefore \;$ The given equation in Cartesian coordinates is $\;$ $2xy = a^2$

Coordinate Geometry - Straight Line

Change to polar coordinates the equation $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$.

The relation between Cartesian coordinates $\left(x, y\right)$ and polar coordinates $\left(r, \theta\right)$ is

$x = r \cos \theta$, $\;\;$ $y = r \sin \theta$

$\begin{aligned} \therefore \; \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 & \implies \dfrac{r^2 \cos^2 \theta}{a^2} + \dfrac{r^2 \sin^2 \theta}{b^2} = 1 \\\\ & i.e. \;\; r^2 \left(\dfrac{b^2 \cos^2 \theta + a^2 \sin^2 \theta}{a^2 b^2}\right) = 1 \\\\ & i.e. \;\; b^2 \cos^2 \theta + a^2 \sin^2 \theta = \dfrac{a^2 b^2}{r^2} \end{aligned}$

$\therefore \;$ The polar form of the given Cartesian equation $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;$ is $\;$ $b^2 \cos^2 \theta + a^2 \sin^2 \theta = \dfrac{a^2 b^2}{r^2}$.

Coordinate Geometry - Straight Line

Change to polar coordinates the equation $\;$ $x^2 + y^2 = a^2$.

The relation between Cartesian coordinates $\left(x, y\right)$ and polar coordinates $\left(r, \theta\right)$ is

$x = r \cos \theta$, $\;\;$ $y = r \sin \theta$

$\begin{aligned} \therefore \; x^2 + y^2 = a^2 & \implies r^2 \cos^2 \theta + r^2 \sin^2 \theta = a^2 \\\\ & i.e. \;\; r^2 \left(\cos^2 \theta + \sin^2 \theta\right) = a^2 \\\\ & i.e. \;\; r^2 = a^2 \\\\ & i.e. \;\; r = a \end{aligned}$

$\therefore \;$ The polar form of the given Cartesian equation $\;$ $x^2 + y^2 = a^2$ $\;$ is $\;$ $r = a$.

Coordinate Geometry - Straight Line

Find the area of the triangle whose vertices are $\left(2, \dfrac{\pi}{3}\right)$, $\left(3, \dfrac{\pi}{2}\right)$, $\left(2, \dfrac{5 \pi}{6}\right)$

The given polar coordinates are $\;$ $A\left(r_1, \theta_1\right) = \left(2, \dfrac{\pi}{3}\right)$, $\;\;$ $B\left(r_2, \theta_2\right) = \left(3, \dfrac{\pi}{2}\right)$, $\;\;$ $C \left(r_3, \theta_3\right) = \left(2, \dfrac{5 \pi}{6}\right)$

Area of triangle $ABC$ is

$\begin{aligned} = \Delta & = \dfrac{r_1 r_2 r_3}{2} \left\{\dfrac{\sin \left(\theta_3 - \theta_2\right)}{r_1} + \dfrac{\sin \left(\theta_1 - \theta_3\right)}{r_2} + \dfrac{\sin \left(\theta_2 - \theta_1\right)}{r_3} \right\} \\\\ & = \dfrac{2 \times 3 \times 2}{2} \left\{\dfrac{\sin \left(\dfrac{5 \pi}{6} - \dfrac{\pi}{2}\right)}{2} + \dfrac{\sin \left(\dfrac{\pi}{3} - \dfrac{5 \pi}{6}\right)}{3} + \dfrac{\sin \left(\dfrac{\pi}{2} - \dfrac{\pi}{3}\right)}{2} \right\} \\\\ & = 6 \times \left\{\dfrac{\sin \left(\dfrac{\pi}{3}\right)}{2} + \dfrac{\sin \left(\dfrac{-\pi}{2}\right)}{3} + \dfrac{\sin \left(\dfrac{\pi}{6}\right)}{2} \right\} \\\\ & = 6 \times \left\{\dfrac{\dfrac{\sqrt{3}}{2}}{2} - \dfrac{1}{3} + \dfrac{\dfrac{1}{2}}{2} \right\} \\\\ & = 6 \times \left\{\dfrac{\sqrt{3}}{4} - \dfrac{1}{3} + \dfrac{1}{4} \right\} \\\\ & = \dfrac{3 \sqrt{3} - 1}{2} \;\; \text{ square units} \end{aligned}$

Coordinate Geometry - Straight Line

Find the distance between the points whose polar coordinates are $\left(a, \dfrac{\pi}{2}\right)$ and $\left(3a, \dfrac{\pi}{6}\right)$.

The given polar coordinates are $\;$ $A\left(r_1, \theta_1\right) = \left(a, \dfrac{\pi}{2}\right)$, $\;\;$ $B\left(r_2, \theta_2\right) = \left(3a, \dfrac{\pi}{6}\right)$

$\begin{aligned} \text{Distance } AB & = \sqrt{r_1^2 + r_2^2 -2 r_1 r_2 \cos \left(\theta_1 - \theta_2\right)} \\\\ & = \sqrt{a^2 + \left(3a\right)^2 - 2 \times a \times 3a \cos \left(\dfrac{\pi}{2} - \dfrac{\pi}{6}\right)} \\\\ & = \sqrt{10a^2 - 6a^2 \cos \left(\dfrac{\pi}{3}\right)} \\\\ & = \sqrt{10a^2 - 3a^2} \\\\ & = a \sqrt{7} \; \text{units} \end{aligned}$

Coordinate Geometry - Straight Line

Find the points whose polar coordinates are:

1. $\left(2, 45^\circ\right)$


2. $\left(5, \dfrac{-\pi}{2}\right)$

1. The polar coordinates are $\;$ $\left(r, \theta\right) = \left(2, 45^\circ\right)$
   
   Let the corresponding Cartesian coordinates be $\left(x, y\right)$
   
   Now, $\;$ $x = r \cos \theta = 2 \cos 45^\circ = \dfrac{2}{\sqrt{2}} = \sqrt{2}$
   
   and $\;$ $y = r \sin \theta = 2 \sin 45^\circ = \dfrac{2}{\sqrt{2}} = \sqrt{2}$
   
   $\therefore \;$ The corresponding Cartesian coordinates are $\;$ $\left(x, y\right) = \left(\sqrt{2}, \sqrt{2}\right)$



2. The polar coordinates are $\;$ $\left(r, \theta\right) = \left(5, \dfrac{-\pi}{2}\right)$
   
   Let the corresponding Cartesian coordinates be $\left(x, y\right)$
   
   Now, $\;$ $x = r \cos \theta = 5 \cos \left(\dfrac{-\pi}{2}\right) = 0$
   
   and $\;$ $y = r \sin \theta = 2 \sin \left(\dfrac{-\pi}{2}\right) = -2$
   
   $\therefore \;$ The corresponding Cartesian coordinates are $\;$ $\left(x, y\right) = \left(0, -2\right)$

Coordinate Geometry - Straight Line

Find the area of the triangle, the equations whose sides are $\;$ $y + x = 0$, $\;$ $y = x + 6$ $\;$ and $\;$ $y = 7x + 5$.

The sides of the triangle are

$x + y = 0$ $\;\;\; \cdots \; (1)$,

$y = x + 6$ $\;$ i.e. $\;$ $x - y + 6 = 0$ $\;\;\; \cdots \; (2)$,

$y = 7x + 5$ $\;$ i.e. $\;$ $7x - y + 5 = 0$ $\;\;\; \cdots \; (3)$

Let equations $(1)$ and $(2)$ intersect at point $A$.

Let equations $(2)$ and $(3)$ intersect at point $B$.

Let equations $(3)$ and $(1)$ intersect at point $C$.

Solving equations $(1)$ and $(2)$ simultaneously gives the point of intersection as $\;$ $A \left(x_1, y_1\right) = \left(-3, 3\right)$

Solving equations $(2)$ and $(3)$ simultaneously gives the point of intersection as $\;$ $B \left(x_2, y_2\right) = \left(\dfrac{1}{6}, \dfrac{37}{6}\right)$

Solving equations $(3)$ and $(1)$ simultaneously gives the point of intersection as $\;$ $C \left(x_3, y_3\right) = \left(\dfrac{-5}{8}, \dfrac{5}{8}\right)$

Let $AD$ be the altitude drawn from point $A$ to line given by equation $(3)$.

Then, length of altitude $AD = \left|\dfrac{7 \times \left(-3\right) -1 \times 3 + 5}{\sqrt{7^2 + \left(-1\right)^2}}\right| = \dfrac{19}{\sqrt{50}} = \dfrac{19}{5 \sqrt{2}}$ units

Distance $BC = \sqrt{\left(\dfrac{1}{6}+ \dfrac{5}{8}\right)^2 + \left(\dfrac{37}{6} - \dfrac{5}{8}\right)^2} = \sqrt{\left(\dfrac{19}{24}\right)^2 + \left(\dfrac{133}{24}\right)^2} = \dfrac{\sqrt{18050}}{24} = \dfrac{95 \sqrt{2}}{24}$ units

$\therefore \;$ Area of triangle whose sides are the given equations is

$= \dfrac{1}{2} \times AD \times BC = \dfrac{1}{2} \times \dfrac{19}{5 \sqrt{2}} \times \dfrac{95 \sqrt{2}}{24} = \dfrac{361}{48}$ sq units

Coordinate Geometry - Straight Line

If $\left(x_1, y_1\right)$ be the coordinates of the foot of the perpendicular from the origin to the line $\dfrac{x}{a} + \dfrac{y}{b} = 1$, show that $\left(x_1^2 + y_1^2\right) \left(x_1 + y_1\right) = \left(a + b\right) x_1 y_1$

Let $A = \left(0, 0\right)$; $\;$ $B = \left(x_1, y_1\right)$ $\;$ be the foot of the perpendicular

Slope of perpendicular $AB = m_1 = \dfrac{y_1}{x_1}$

Equation of given line is $\;$ $\dfrac{x}{a} + \dfrac{y}{b} = 1$

i.e. $\;$ $bx + ya = ab$

$\therefore \;$ Slope of given line $= m_2 = \dfrac{-b}{a}$

$\because \;$ $AB$ is perpendicular to the given line, $\;$ $m_1 = \dfrac{-1}{m_2}$

i.e. $\;$ $\dfrac{y_1}{x_1} = \dfrac{a}{b}$ $\implies$ $y_1 = \dfrac{a x_1}{b}$ $\;\;\; \cdots \; (1)$

$\because \;$ $B$ is the foot of the perpendicular, point $B$ lies on the given line.

$\therefore \;$ We have, $\;$ $\dfrac{x_1}{a} + \dfrac{y_1}{b} = 1$

i.e. $\;$ $bx_1 + ay_1 = ab$

i.e. $\;$ $bx_1 + a \times \dfrac{ax_1}{b} = ab$ $\;\;\;$ [in view of equation $(1)$]

i.e. $\;$ $b^2 x_1 + a^2 x_1 = ab^2$

i.e. $\;$ $\left(a^2 + b^2\right) x_1 = ab^2$ $\implies$ $a^2 + b^2 = \dfrac{ab^2}{x_1}$ $\;\;\; \cdots \; (2)$

Now,

$\begin{aligned} \left(x_1^2 + y_1^2\right) \left(x_1 + y_1\right) & = \left(x_1^2 + \dfrac{a^2 x_1^2}{b^2}\right) \left(x_1 + \dfrac{ax_1}{b}\right) \\\\ & = \dfrac{\left(b^2 x_1^2 + a^2 x_1^2\right) \left(b x_1 + a x_1\right)}{b^3} \\\\ & = \dfrac{x_1^2 \left(a^2 + b^2\right) x_1 \left(a + b\right)}{b^3} \\\\ & = \dfrac{x_1^3 \left(a^2 + b^2\right) \left(a + b\right)}{b^3} \\\\ & = x_1^3 \times \dfrac{ab^2}{x_1} \times \dfrac{\left(a + b\right)}{b^3} \;\;\; \left[\text{by equation (2)}\right] \\\\ & = \dfrac{a \left(a + b\right)x_1^2}{b} \;\;\; \cdots \; (3) \end{aligned}$

$\begin{aligned} \left(a + b\right) x_1 y_1 & = \left(a + b\right) \times x_1 \times \dfrac{ax_1}{b} \\\\ & = \dfrac{a \left(a + b\right) x_1^2}{b} \;\;\; \cdots \; (4) \end{aligned}$

$\therefore \;$ We have from equations $(3)$ and $(4)$

$\left(x_1^2 + y_1^2\right) \left(x_1 + y_1\right) = \left(a + b\right) x_1 y_1$

Hence proved.

Coordinate Geometry - Straight Line

Find the coordinates of the foot of the perpendicular from the point $\left(-2, 0\right)$ on the line $2x + 3y = 9$.

Given point $= A \left(x_1, y_1\right) = \left(-2, 0\right)$

Let the foot of the perpendicular from point $A$ be at the point $\;$ $B \left(x_2, y_2\right)$

Equation of given line is $\;$ $2x + 3y = 9$ $\;\;\; \cdots \; (1)$

Slope of the given line $\;$ $2x + 3y = 9$ is $\;$ $m = \dfrac{-2}{3}$

$\because \;$ $AB$ is perpendicular to the given line, slope of perpendicular $= m_1 = \dfrac{-1}{m} = \dfrac{3}{2}$

$\therefore \;$ Equation of perpendicular $AB$ passing through the point $A$ and having slope $= m_1$ is

$y - 0 = \dfrac{3}{2} \left(x + 2\right)$

i.e. $\;$ $2y = 3x + 6$

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

Solving equations $(1)$ and $(2)$ simultaneously gives the point of intersection as $\left(0, 3\right)$

The point of intersection is the foot of the perpendicular from point $A$ on the line given by equation $(1)$.

$\therefore \;$ $B = \left(0, 3\right)$

Coordinate Geometry - Straight Line

The equations of two sides of a triangle are $\;$ $2x - y = 0$ $\;$ and $\;$ $x + 3y - 7 = 0$. The orthocenter is the point $\left(2,1\right)$. Find the equation of the third side.

The sides of the triangle are

$2x - y = 0$ $\;\;\; \cdots \; (1)$, $\;$ $x + 3y - 7 = 0$ $\;\;\; \cdots \; (2)$

Let the equation of the third side of the triangle be $\;$ $ax + by + c = 0$ $\;\;\; \cdots \; (3)$

Let equations $(1)$ and $(2)$ intersect at point $A$.

Let equations $(2)$ and $(3)$ intersect at point $B$.

Solving equations $(1)$ and $(2)$ simultaneously gives the point of intersection as $\;$ $A \left(x_1, y_1\right) = \left(1, 2\right)$

Solving equations $(2)$ and $(3)$ simultaneously gives the point of intersection as $\;$ $B \left(x_2, y_2\right) = \left(\dfrac{-7b-3c}{3a-b}, \dfrac{7a+c}{3a-b}\right)$

Let $AD$ be the altitude drawn from point $A$ to line given by equation $(3)$.

Slope of equation $(3)$ is $m_1 = \dfrac{-a}{b}$

$\therefore \;$ Slope of altitude $AD = \dfrac{-1}{m_1} = \dfrac{b}{a}$

$AD$ passes through the point $A$ and has slope $\dfrac{-1}{m_1}$.

$\therefore \;$ Equation of $AD$ is

$y - 2 = \dfrac{b}{a} \left(x - 1\right)$

i.e. $\;$ $ay - 2a= bx - b$

i.e. $\;$ $bx - ay + 2a-b = 0$ $\;\;\; \cdots \; (4)$

Let $BE$ be the altitude drawn from point $B$ to line given by equation $(1)$.

Slope of equation $(1)$ is $m_2 = 2$

$\therefore \;$ Slope of altitude $BE = \dfrac{-1}{m_2} = \dfrac{-1}{2}$

$BE$ passes through the point $B$ and has slope $\dfrac{-1}{m_2}$.

$\therefore \;$ Equation of $BE$ is

$y - \left(\dfrac{7a + c}{3a - b}\right) = \dfrac{-1}{2} \left[x + \dfrac{7b + 3c}{3a - b}\right]$

i.e. $\;$ $\left(3a - b\right)y - 7a - c = \dfrac{-1}{2} \left[\left(3a - b\right)x + 7b + 3c\right]$

i.e. $\;$ $\left(3a - b\right)x + \left(6a - 2b\right)y - 14a + 7b + c = 0$ $\;\;\; \cdots \; (5)$

Given: Orthocenter of the triangle is $\;$ $O \left(2, 1\right)$.

$\therefore \;$ Point $O$ satisfies both equations $(4)$ and $(5)$.

Substituting $\left(x, y\right) = \left(2,1\right)$ in equation $(4)$ gives,

$2b - a + 2a - b = 0$

i.e. $\;$ $a + b = 0$ $\implies$ $a = -b$ $\;\;\; \cdots \; (6)$

Substituting $\left(x, y\right) = \left(2,1\right)$ in equation $(5)$ gives,

$2 \left(3a - b\right) + 6a - 2b - 14a + 7b + c = 0$

i.e. $\;$ $6a - 2b + 6a - 2b - 14a + 7b + c = 0$

i.e. $\;$ $-2a + 3b + c = 0$ $\;\;\; \cdots \; (7)$

In view of equation $(6)$, equation $(7)$ becomes

$2b + 3b + c = 0$ $\implies$ $c = -5b$ $\;\;\; \cdots \; (8)$

$\therefore \;$ In view of equations $(6)$ and $(8)$ equation $(3)$ becomes,

$-bx + by -5b = 0$

$\therefore \;$ The equation of the third side of the triangle is

$x - y + 5 = 0$

Coordinate Geometry - Straight Line

The equations of the sides of a triangle are $\;$ $x - y - 1 = 0$, $\;$ $2x - 5 = 0$ $\;$ and $\;$ $3x + y -2 = 0$. Find the coordinates of the orthocenter.

The sides of the triangle are

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

Let equations $(1)$ and $(2)$ intersect at point $A$.

Let equations $(2)$ and $(3)$ intersect at point $B$.

Let equations $(3)$ and $(1)$ intersect at point $C$.

Solving equations $(1)$ and $(2)$ simultaneously gives the point of intersection as $\;$ $A \left(x_1, y_1\right) = \left(\dfrac{5}{2}, \dfrac{3}{2}\right)$

Solving equations $(2)$ and $(3)$ simultaneously gives the point of intersection as $\;$ $B \left(x_2, y_2\right) = \left(\dfrac{5}{2}, \dfrac{-11}{2}\right)$

Solving equations $(3)$ and $(1)$ simultaneously gives the point of intersection as $\;$ $C \left(x_3, y_3\right) = \left(\dfrac{3}{4}, \dfrac{-1}{4}\right)$

Let $AD$ be the altitude drawn from point $A$ to line given by equation $(3)$.

Slope of equation $(3)$ is $m_1 = -3$

$\therefore \;$ Slope of altitude $AD = \dfrac{-1}{m_1} = \dfrac{1}{3}$

$AD$ passes through the point $A$ and has slope $\dfrac{-1}{m_1}$.

$\therefore \;$ Equation of $AD$ is

$y - \dfrac{3}{2} = \dfrac{1}{3} \left(x - \dfrac{5}{2}\right)$

i.e. $\;$ $6y - 9 = 2x - 5$

i.e. $\;$ $x - 3y + 2 = 0$ $\;\;\; \cdots \; (4)$

Let $BE$ be the altitude drawn from point $B$ to line given by equation $(1)$.

Slope of equation $(1)$ is $m_2 = 1$

$\therefore \;$ Slope of altitude $BE = \dfrac{-1}{m_2} = -1$

$BE$ passes through the point $B$ and has slope $\dfrac{-1}{m_2}$.

$\therefore \;$ Equation of $BE$ is

$y + \dfrac{11}{2} = -1 \left(x - \dfrac{5}{2}\right)$

i.e. $\;$ $2y + 11 = -2x + 5$

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

Solving equations $(4)$ and $(5)$ simultaneously gives the point of intersection as $\left(\dfrac{-11}{4}, \dfrac{-1}{4}\right)$

$\therefore \;$ The orthocenter is $\left(\dfrac{-11}{4}, \dfrac{-1}{4}\right)$.