Analytical Geometry - Conics - Hyperbola

Find the eccentricity, center, foci and vertices of the hyperbola $25x^2 - 16y^2 = 400$ and sketch it.

Equation of given hyperbola is: $\;$ $25x^2 - 16y^2 = 400$

i.e. $\;$ $\dfrac{x^2}{400 / 25} - \dfrac{y^2}{400 / 16} = 1$

i.e. $\;$ $\dfrac{x^2}{16} - \dfrac{y^2}{25} = 1$ $\;\;\; \cdots \; (1)$

The transverse axis is along the X axis.

Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 16 \implies a = 4$; $\;$ and $\;$ $b^2 = 25 \implies b = 5$

Now, eccentricity $= e = \sqrt{1 + \dfrac{b^2}{a^2}} = \sqrt{1 + \dfrac{25}{16}} = \dfrac{\sqrt{41}}{4}$

Center $= C \left(0, 0\right)$

Foci [$F_1$ and $F_2$] $= \left(\pm ae, 0\right) = \left(\pm 4 \times \dfrac{\sqrt{41}}{4}, 0\right) = \left(\pm \sqrt{41}, 0\right)$

Vertices [$A_1$ and $A_2$] $= \left(\pm a, 0\right) = \left(\pm 4, 0\right)$

Analytical Geometry - Conics - Hyperbola

Find the equations of directrices, latus rectums and length of latus rectum for the hyperbola $9x^2 - 4y^2 - 36x + 32y + 8 = 0$

Equation of given hyperbola is: $\;$ $9x^2 - 4y^2 - 36x + 32y + 8 = 0$

i.e. $\;$ $9 \left(x^2 - 4x\right) - 4 \left(y^2 - 8y\right) = -8$

i.e. $\;$ $9 \left[\left(x^2 - 4x + 4\right) - 4\right] - 4 \left[\left(y^2 - 8y + 16\right) - 16\right] = -8$

i.e. $\;$ $9 \left(x - 2\right)^2 - 4 \left(y - 4\right)^2 = 36 - 64 - 8$

i.e. $\;$ $9 \left(x - 2\right)^2 - 4 \left(y - 4\right)^2 = -36$

i.e. $\;$ $\dfrac{\left(y - 4\right)^2}{36 /4} - \dfrac{\left(x - 2\right)^2}{36 / 9} = 1$

i.e. $\;$ $\dfrac{\left(y - 4\right)^2}{9} - \dfrac{\left(x - 2\right)^2}{4} = 1$ $\;\;\; \cdots \; (1)$

Let $Y = y - 4$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $X = x - 2$ $\;\;\; \cdots \; (2b)$

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

$\dfrac{Y^2}{9} - \dfrac{X^2}{4} = 1$ $\;\;\; \cdots \; (3)$

For the hyperbola given by equation $(3)$, the transverse axis is along the Y axis.

Comparing equation $(3)$ with the standard equation of hyperbola $\;$ $\dfrac{Y^2}{a^2} - \dfrac{X^2}{b^2} = 1$ $\;$ gives

$a^2 = 9 \implies a = 3$ $\;$ and $\;$ $b^2 = 4 \implies b = 2$

Now, $e = \sqrt{1 + \dfrac{b^2}{a^2}} = \sqrt{1 + \dfrac{4}{9}} = \dfrac{\sqrt{13}}{3}$

	Referred to X, Y	Referred to x, y $X = x - 2$; $\;$ $Y = y - 4$ i.e. $\;$ $x = X + 2$; $\;$ $y = Y + 4$
Equations of directrices	$Y = \pm \dfrac{a}{e} = \pm \dfrac{3}{\sqrt{13} / 3}$ i.e. $\;$ $Y = \pm \dfrac{9}{\sqrt{13}}$	$Y = \pm \dfrac{9}{\sqrt{13}}$ $ \implies y = 4 \pm \dfrac{9}{\sqrt{13}}$
Equations of latus rectum	$Y = \pm a \cdot e = \pm 3 \times \dfrac{\sqrt{13}}{3}$ i.e. $\;$ $Y = \pm \sqrt{13}$	$Y = \pm \sqrt{13}$ $\implies$ $y = 4 \pm \sqrt{13}$
Length of latus rectum	$\dfrac{2b^2}{a} = \dfrac{2 \times 4}{3} = \dfrac{8}{3}$	$\dfrac{8}{3}$

Analytical Geometry - Conics - Hyperbola

Find the equations and length of transverse and conjugate axes of the hyperbola $\;$ $16x^2 - 9y^2 + 96x + 36y - 36 = 0$

Equation of given hyperbola is: $\;$ $16x^2 - 9y^2 + 96 x + 36 y - 36 = 0$

i.e. $\;$ $16 \left(x^2 + 6x\right) - 9 \left(y^2 - 4y\right) = 36$

i.e. $\;$ $16 \left[\left(x^2 + 6x + 9\right) - 9\right] - 9 \left[\left(y^2 - 4y + 4\right) - 4\right] = 36$

i.e. $\;$ $16 \left(x + 3\right)^2 - 9 \left(y - 2\right)^2 = 144 - 36 + 36$

i.e. $\;$ $16 \left(x + 3\right)^2 - 9 \left(y - 2\right)^2 = 144$

i.e. $\;$ $\dfrac{\left(x + 3\right)^2}{144 / 16} - \dfrac{\left(y - 2\right)^2}{144 / 9} = 1$

i.e. $\;$ $\dfrac{\left(x + 3\right)^2}{9} - \dfrac{\left(y - 2\right)^2}{16} = 1$ $\;\;\; \cdots \; (1)$

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

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

$\dfrac{X^2}{9} - \dfrac{Y^2}{16} = 1$ $\;\;\; \cdots \; (3)$

The transverse axis of the hyperbola given by equation $(3)$ is along the X axis.

Comparing equation $(3)$ with the standard equation of hyperbola $\;$ $\dfrac{X^2}{a^2} - \dfrac{Y^2}{b^2} = 1$ $\;$ gives

$a^2 = 9 \implies a = 3$; $\;$ $b^2 = 16 \implies b = 4$

	Referred to X, Y	Referred to x, y $X = x + 3$; $\;$ $Y = y - 2$
Equation of transverse axis	X axis $\;$ $\left(Y = 0\right)$	$Y = 0 \implies y - 2 = 0$
Equation of conjugate axis	Y axis $\;$ $\left(X = 0\right)$	$X = 0 \implies x + 3 = 0$
Length of transverse axis	$2a = 2 \times 3 = 6$	$6$
Length of conjugate axis	$2b = 2 \times 4 = 8$	$8$

Analytical Geometry - Conics - Hyperbola

Find the equation of the hyperbola if the center is $\left(1, 4\right)$; one of the foci is $\left(6, 4\right)$ and the corresponding directrix is $x = \dfrac{9}{4}$

Given: Directrix of hyperbola is: $\;$ $x = \dfrac{9}{4}$

$\implies$ the directrix is parallel to the Y axis.

$\therefore$ $\;$ Let the equation of the required hyperbola be:

$\dfrac{\left(x - h\right)^2}{a^2} - \dfrac{\left(y - k\right)^2}{b^2} = 1$ $\;\;\; \cdots \; (1)$

Given: Center $C \left(h, k\right) = \left(1, 4\right)$ $\;\;\; \cdots \; (2a)$; Focus $F_1 = \left(6, 4\right)$ $\;\;\; \cdots \; (2b)$

$CF_1 = a \cdot e = \sqrt{\left(6 - 1\right)^2 + \left(4 - 4\right)^2}$

i.e. $\;$ $a \cdot e = 5$ $\;\;\; \cdots \; (3)$

Draw CZ perpendicular to the directrix.

Then, $Z = \left(\dfrac{9}{4}, 4\right)$

Distance between center and directrix is $= CZ = \dfrac{a}{e} = \sqrt{\left(\dfrac{9}{4} - 1\right)^2 + \left(4 - 4\right)^2}$

i.e. $\;$ $\dfrac{a}{e} = \dfrac{5}{4}$ $\;\;\; \cdots \; (4)$

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

$a \cdot e \times \dfrac{a}{e} = 5 \times \dfrac{5}{4}$

i.e. $\;$ $a^2 = \dfrac{25}{4}$ $\;\;\; \cdots \; (5a)$

and $\dfrac{a \cdot e}{a / e} = \dfrac{5}{5 / 4}$ $\implies$ $e^2 = 4$

Now, $b^2 = a^2 \times \left(e^2 - 1\right) = \dfrac{25}{4} \times \left(4 - 1\right)$

i.e. $\;$ $b^2 = \dfrac{75}{4}$ $\;\;\; \cdots \; (5b)$

$\therefore$ $\;$ In view of equations $(2a)$, $(5a)$ and $(5b)$, the equation of required hyperbola [equation $(1)$] becomes

$\dfrac{\left(x -1 \right)^2}{25 / 4} - \dfrac{\left(y - 4\right)^2}{75 / 4} = 1$

Analytical Geometry - Conics - Hyperbola

Find the equation of the hyperbola if foci are $\left(\pm 3, 5\right)$ and $e = 3$.

Given: Foci $F_1 = \left(3, 5\right)$ and $F_2 = \left(-3, 5\right)$; $\;$ $e = 3$

Distance between foci $= F_1 F_2 = \sqrt{\left(3 + 3\right)^2 + \left(5 - 5\right)^2}$

i.e. $\;$ $F_1 F_2 = 6$ $\;\;\; \cdots \; (1)$

But $F_1 F_2 = 2 a e$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ We have from equations $(1)$ and $(2)$,

$2a \times 3 = 6$ $\implies$ $a = 1$

Now, $b^2 = a^2 \left(e^2 - 1\right) = 1 \times \left(9 - 1\right) = 8$

The foci are $\left(\pm 3, 5\right)$ $\implies$ the transverse axis is parallel to the X axis.

$\therefore$ $\;$ Let the equation of the required hyperbola be:

$\dfrac{\left(x - h\right)^2}{a^2} - \dfrac{\left(y - k\right)^2}{b^2} = 1$ $\;\;\; \cdots \; (3)$

Center $C \left(h, k\right)$ is the midpoint of $F_1$ and $F_2$.

$\therefore$ $\;$ $C\left(h, k\right) = \left(\dfrac{3 - 3}{2}, \dfrac{5 + 5}{2}\right) = \left(0, 5\right)$

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

$\dfrac{\left(x - 0\right)^2}{1^2} - \dfrac{\left(y - 5\right)^2}{8} = 1$

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

Analytical Geometry - Conics - Hyperbola

Find the equation of the hyperbola if the center is $\left(1, -2\right)$; length of transverse axis is 8; $e = \dfrac{5}{4}$ and the transverse axis parallel to the X axis.

Given: Center $C \left(h, k\right) = \left(1, -2\right)$; $\;$ $e = \dfrac{5}{4}$

The transverse axis is parallel to the X axis.

$\therefore$ $\;$ Let the equation of the required hyperbola be: $\;$ $\dfrac{\left(x - h\right)^2}{a^2} - \dfrac{\left(y - k\right)^2}{b^2} = 1$

Length of transverse axis $= 2a = 8$ $\implies$ $a = 4$

Now, $b^2 = a^2 \left(e^2 - 1\right)$

i.e. $\;$ $b^2 = 16 \times \left(\dfrac{25}{16} - 1\right) = 16 \times \dfrac{9}{16} = 9$

$\therefore$ $\;$ Equation of required hyperbola is: $\;$ $\dfrac{\left(x -1\right)^2}{16} - \dfrac{\left(y + 2\right)^2}{9} = 1$

Analytical Geometry - Conics - Hyperbola

Find the equation of the hyperbola if the focus is $\left(2,3\right)$; the corresponding directrix is $x + 2y = 5$ and $e = 2$.

Given: Equation of directrix is $\;$ $x + 2y = 5$; $\;$ $e = 2$

Let $P \left(x, y\right)$ be any point on the hyperbola.

Draw $PM$ perpendicular to the directrix.

By definition, $\dfrac{FP}{PM} = e$

i.e. $\;$ $FP^2 = e^2 \cdot PM^2$

i.e. $\;$ $\left(x - 2\right)^2 + \left(y - 3\right)^2 = \left(2\right)^2 \times \left(\dfrac{x + 2y - 5}{\sqrt{1^2 + 2^2}}\right)^2$

i.e. $\;$ $x^2 - 4x + 4 + y^2 - 6y + 9 = \dfrac{4}{5} \left(x^2 + 4y^2 + 25 + 4xy - 10x - 20y\right)$

i.e. $\;$ $5x^2 - 20x + 5y^2 - 30y + 65 = 4x^2 + 16y^2 + 100 + 16xy - 40x - 80y$

i.e. $\;$ $x^2 - 16xy - 11y^2 + 20x + 50y - 35 = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is the required equation of hyperbola.

Analytical Geometry - Conics - Ellipse

The orbit of the planet mercury around the sun is in an elliptical shape with the sun at a focus. The semi-major axis is of length 36 million miles and the eccentricity of the orbit is 0.206. Find

1. how close does mercury get to the sun;
2. the greatest possible distance between mercury and sun.

Let the equation of the elliptical orbit be: $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

Center of the ellipse is at $C \left(0,0\right)$.

Let the sun be at focus $F_1$.

Vertices of the ellipse are $A_1$ and $A_2$.

Semi-major axis of the ellipse $= CA_1 = CA_2 = a = 36$ million miles

Eccentricity of the elliptical orbit $= e = 0.206$

Now, $CF_1 = a \times e = 36 \times 0.206 = 7.416$ million miles

1. When mercury is at point $A_1$, it is closest to the sun.
   
   $\therefore$ $\;$ Required distance $= F_1 A_1$
   
   Now, $F_1 A_1 = CA_1 - CF_1 = 36 - 7.416 = 28.584$ million miles
   
   
2. When mercury is at the point $A_2$, it is farthest from the sun.
   
   $\therefore$ $\;$ Required distance $= F_1 A_2$
   
   Now, $F_1 A_2 = CF_1 + CA_2 = 7.416 + 36 = 43.416$ million miles

Analytical Geometry - Conics - Ellipse

Find the eccentricity, center, foci and vertices of the ellipse $\;$ $9x^2 + 4y^2 = 36$ $\;$ and sketch it.

Equation of given ellipse is: $\;$ $9x^2 + 4y^2 = 36$

i.e. $\;$ $\dfrac{x^2}{36/9} + \dfrac{y^2}{36/4} = 1$

i.e. $\;$ $\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$ $\;\;\; \cdots \; (1)$

Major axis of the ellipse is along the Y axis.

Comparing equation $(1)$ with the standard equation of the ellipse $\;$ $\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1$ $\;$ gives

$b^2 = 4 \implies b = 2$; $\;$ and $\;$ $a^2 = 9 \implies a = 3$

Now, $e = \sqrt{1 - \dfrac{b^2}{a^2}} = \sqrt{1 - \dfrac{4}{9}} = \dfrac{\sqrt{5}}{3}$

$\therefore$ $\;$ $a \times e = 3 \times \dfrac{\sqrt{5}}{3} = \sqrt{5}$

Center $= \left(0, 0\right)$

Foci $\left(F_1 \text{ and } F_2\right) = \left(0, \pm ae\right) = \left(0, \pm \sqrt{5}\right)$

Vertices $\left(A_1 \text{ and } A_2\right) = \left(0, \pm a\right) = \left(0, \pm 3\right)$

Analytical Geometry - Conics - Ellipse

Find the eccentricity, center, foci and vertices of the ellipse $\;$ $x^2 + 4y^2 -8x - 16 y - 68 = 0$ $\;$ and sketch it.

Equation of the given ellipse is: $\;$ $x^2 + 4y^2 - 8x - 16 y - 68 = 0$

i.e. $\;$ $\left(x^2 - 8x\right) + 4 \left(y^2 - 4y\right) = 68$

i.e. $\;$ $\left[\left(x^2 - 8x + 16\right) - 16\right] + 4 \left[\left(y^2 - 4y + 4\right) - 4\right] = 68$

i.e. $\;$ $\left(x - 4\right)^2 + 4 \left(y - 2\right)^2 = 68 + 16 + 16$

i.e. $\;$ $\left(x - 4\right)^2 + 4 \left(y - 2\right)^2 = 100$

i.e. $\;$ $\dfrac{\left(x - 4\right)^2}{100} + \dfrac{\left(y - 2\right)^2}{100 / 4} = 1$

i.e. $\;$ $\dfrac{\left(x - 4\right)^2}{100} + \dfrac{\left(y - 2\right)^2}{25} = 1$ $\;\;\; \cdots \; (1)$

Let $X = x - 4$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $Y = y - 2$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2b)$, equation $(1)$ can be written as: $\;$ $\dfrac{X^2}{100} + \dfrac{Y^2}{25} = 1$ $\;\;\; \cdots \; (3)$

Major axis of the ellipse given by equation $(3)$ is along the X axis.

Comparing equation $(3)$ with the standard equation of ellipse $\;$ $\dfrac{X^2}{a^2} + \dfrac{Y^2}{b^2} = 1$ $\;$ gives

$a^2 = 100$ $\implies$ $a = 10$; $\;\;$ $b^2 = 25$ $\implies$ $b = 5$

$\therefore$ $\;$ Eccentricity $= e = \sqrt{1 - \dfrac{b^2}{a^2}} = \sqrt{1 - \dfrac{25}{100}} = \sqrt{\dfrac{75}{100}} = \dfrac{\sqrt{3}}{2}$

Now, $a \times e = 10 \times \dfrac{\sqrt{3}}{2} = 5 \sqrt{3}$

	Referred to X, Y	Referred to x, y $X = x - 4$; $\;$ $Y = y - 2$ i.e. $\;$ $x = X + 4$; $\;$ $y = Y + 2$
Center	$\left(0, 0\right)$	$X = 0 \implies x = 4$ $Y = 0 \implies y = 2$ $\therefore$ $\;$ Center $= \left(4, 2\right)$
Foci	$F_1 = \left(ae, 0\right)$ $F_1 = \left(10 \times \dfrac{\sqrt{3}}{2}, 0\right)$ i.e. $\;$ $F_1 = \left(5 \sqrt{3}, 0\right)$	$X = 5 \sqrt{3} \implies x = 5 \sqrt{3} + 4$ $Y = 0 \implies y = 2$ $\therefore$ $\;$ $F'_1 = \left(4 + 5 \sqrt{3}, 2\right)$
	$F_2 = \left(-ae, 0\right)$ i.e. $\;$ $F_2 = \left(-10 \times \dfrac{\sqrt{3}}{2}, 0\right)$ i.e. $\;$ $F_2 = \left(-5 \sqrt{3}, 0\right)$	$X = -5 \sqrt{3} \implies x = -5 \sqrt{3} + 4$ $Y = 0 \implies y = 2$ $\therefore$ $\;$ $F'_2 = \left(4 - 5 \sqrt{3}, 2\right)$
Vertices	$A_1 = \left(+a, 0\right)$ i.e. $\;$ $A_1 = \left(10, 0\right)$	$X = 10 \implies x = 14$ $Y = 0 \implies y = 2$ $\therefore$ $\;$ $A'_1 = \left(14, 2\right)$
	$A_2 = \left(-a, 0\right)$ i.e. $\;$ $A_2 = \left(-10, 0\right)$	$X = -10 \implies x = -6$ $Y = 0 \implies y = 2$ $\therefore$ $\;$ $A'_2 = \left(-6, 2\right)$

Analytical Geometry - Conics - Ellipse

Find the equations of directrices, latus rectum and length of latus rectum for the ellipse $3x^2 + 2y^2 - 30 x - 4y + 23 = 0$

Equation of given ellipse is: $\;$ $3x^2 + 2y^2 - 30 x - 4y + 23 = 0$

i.e. $\;$ $3 \left(x^2 - 10x\right) + 2 \left(y^2 - 2y\right) = -23$

i.e. $\;$ $3 \left[\left(x^2 - 10x + 25\right) - 25\right] + 2 \left[\left(y^2 - 2y + 1\right) -1\right] = -23$

i.e. $\;$ $3 \left(x - 5\right)^2 + 2 \left(y - 1\right)^2 - 75 - 2 = - 23$

i.e. $\;$ $3 \left(x - 5\right)^2 + 2 \left(y - 1\right)^2 = 54$

i.e. $\;$ $\dfrac{\left(x - 5\right)^2}{54 / 3} + \dfrac{\left(y - 1\right)^2}{54 / 2} = 1$

i.e. $\;$ $\dfrac{\left(x - 5\right)^2}{18} + \dfrac{\left(y - 1\right)^2}{27} = 1$ $\;\;\; \cdots \; (1)$

Let $\;$ $X = x - 5$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $Y = y - 1$ $\;\;\; \cdots \; (2b)$

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

$\dfrac{X^2}{18} + \dfrac{Y^2}{27} = 1$ $\;\;\; \cdots \; (3)$

Comparing equation $(3)$ with the standard equation of ellipse $\;$ $\dfrac{X^2}{b^2} + \dfrac{Y^2}{a^2} = 1$ $\;$ (for $b < a$) $\;$ gives

$b^2 = 18$ $\;\;\; \cdots \; (4a)$; $\;$ $a^2 = 27 \implies a = 3 \sqrt{3}$ $\;\;\; \cdots \; (4b)$

Now, $e^2 = 1 - \dfrac{b^2}{a^2} = 1 - \dfrac{18}{27} = \dfrac{9}{27}$ $\implies$ $e = \dfrac{1}{\sqrt{3}}$ $\;\;\; \cdots \; (4c)$

Equations of directrices are $\;$ $Y = \pm \dfrac{a}{e}$ $\;\;\; \cdots \; (5)$

$\therefore$ $\;$ In view of equations $(2b)$, $(4b)$ and $(4c)$, equation $(5)$ becomes

$y - 1 = \pm \dfrac{3 \sqrt{3}}{3}$

i.e. $\;$ $y - 1 = \pm 9$

i.e. $\;$ $y = 10$ $\;\;\; \cdots \; (5a)$ $\;$ and $\;$ $y = -8$ $\;\;\; \cdots \; (5b)$

Equations of latus rectum are: $\;$ $Y = \pm ae$ $\;\;\; \cdots \; (6)$

$\therefore$ $\;$ In view of equations $(2b)$, $(4b)$ and $(4c)$, equation $(6)$ becomes

$y - 1 = \pm 3 \sqrt{3 \times \dfrac{1}{\sqrt{3}}}$

i.e. $\;$ $y - 1 = \pm 3$

i.e. $\;$ $y = 4$ $\;\;\; \cdots \; (6a)$ $\;$ and $y = -2$ $\;\;\; \cdots \; (6b)$

Length of latus rectum $= \dfrac{2b^2}{a} = \dfrac{2 \times 18}{3 \sqrt{3}} = 4 \sqrt{3}$ $\;$ [from equations $(4a)$ and $(4b)$]

Analytical Geometry - Conics - Ellipse

Find the equations of directrices, latus rectum and length of latus rectum for the ellipse $9x^2 + 16y^2 = 144$

Equation of the given ellipse is: $\;\;$ $9x^2 + 16y^2 = 144$

i.e. $\;$ $\dfrac{x^2}{144 / 9} + \dfrac{y^2}{144 / 16} = 1$

i.e. $\;$ $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of ellipse $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 16 \implies a = 4$; $\;$ $b^2 = 9$

Now, $\;$ $e^2 = 1 - \dfrac{b^2}{a^2} = 1 - \dfrac{9}{16} = \dfrac{7}{16}$ $\implies$ $e = \dfrac{\sqrt{7}}{4}$

Equations of directrices are: $\;$ $x = \pm \dfrac{a}{e}$

i.e. $\;$ $x = \pm \dfrac{4}{\sqrt{7} / 4}$

i.e. $\;$ $x = \pm \dfrac{16}{\sqrt{7}}$

Equations of latus rectum are: $\;$ $x = \pm ae$

i.e. $\;$ $x = \pm 4 \times \dfrac{\sqrt{7}}{4}$

i.e. $\;$ $x = \pm \sqrt{7}$

Length of latus rectum $= \dfrac{2b^2}{a} = 2 \times \dfrac{9}{4} = \dfrac{9}{2}$

Analytical Geometry - Conics - Ellipse

Find the equation and length of major and minor axes of $\;$ $9x^2 + 4y^2 = 20$

Equation of the given ellipse is: $\;$ $9x^2 + 4y^2 = 20$

i.e. $\;$ $\dfrac{x^2}{20 / 9} + \dfrac{y^2}{20 / 4} =1$

i.e. $\;$ $\dfrac{x^2}{20 / 9} + \dfrac{y^2}{5} = 1$ $\;\;\; \cdots \; (1)$

The major axis of the given ellipse is along the Y axis and the minor axis is along the X axis.

$\therefore$ $\;$ Equation of major axis is: $\;$ $x = 0$

Equation of minor axis is: $\;$ $y = 0$

Comparing equation $(1)$ with the standard equation of ellipse $\;$ $\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1$ $\;$ $\text{for } \left(b < a\right)$ $\;$ gives

$a^2 = 5 \implies a = \sqrt 5$ $\;$ and $\;$ $b^2 = \dfrac{20}{9} \implies b = \dfrac{2 \sqrt{5}}{3}$

$\therefore$ $\;$ Length of major axis $= 2a = 2 \sqrt{5}$

Length of minor axis $= 2b = 2 \times \dfrac{2 \sqrt{5}}{3} = \dfrac{4 \sqrt{5}}{3}$

Analytical Geometry - Conics - Ellipse

Find the equation and length of major and minor axes of $\;$ $5x^2 + 9y^2 + 10 x - 36 y - 4 = 0$

Equation of the given ellipse is: $\;$ $5x^2 + 9y^2 + 10 x - 36 y - 4 = 0$

i.e. $\;$ $\left(5x^2 + 10x\right) + \left(9y^2 - 36y\right) = 4$

i.e. $\;$ $5 \left(x^2 + 2x\right) + 9 \left(y^2 - 4y\right) = 4$

i.e. $\;$ $5 \left[\left(x^2 + 2x + 1\right) - 1\right] + 9 \left[\left(y^2 - 4y + 4\right) -4\right] = 4$

i.e. $\;$ $5 \left[\left(x + 1\right)^2 - 1\right] + 9 \left[\left(y - 2\right)^2 - 4\right] = 4$

i.e. $\;$ $5 \left(x + 1\right)^2 + 9 \left(y - 2\right)^2 = 4 + 5 + 36$

i.e. $\;$ $5 \left(x + 1\right)^2 + 9 \left(y - 2\right)^2 = 45$

i.e. $\;$ $\dfrac{\left(x + 1\right)^2}{45 / 5} + \dfrac{\left(y - 2\right)^2}{45 / 9} = 1$

i.e. $\;$ $\dfrac{\left(x + 1\right)^2}{9} + \dfrac{\left(y -2 \right)^2}{5} = 1$ $\;\;\; \cdots \; (1)$

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

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

$\dfrac{X^2}{9} + \dfrac{Y^2}{5} = 1$ $\;\;\; \cdots \; (3)$

The major axis is along the X axis and the minor axis is along the Y axis.

$\therefore$ $\;$ Equation of major axis is: $\;$ $Y = 0$ $\implies$ $y - 2 = 0$

Equation of minor axis is: $\;$ $X = 0$ $\implies$ $x + 1 = 0$

Comparing equation $(3)$ with the standard equation of ellipse $\;$ $\dfrac{X^2}{a^2} + \dfrac{Y^2}{b^2} = 1$ $\;$ $\left(\text{for } a > b\right)$ $\;$ gives

$a^2 = 9 \implies a = 3$; $\;$ and $\;$ $b^2 = 5 \implies b = \sqrt{5}$

$\therefore$ $\;$ Length of major axis $= 2a = 2 \times 3 = 6$

Length of minor axis $= 2b = 2 \sqrt{5}$

Analytical Geometry - Conics - Ellipse

Find the locus of a point which moves so that the sum of its distances from $\left(3,0\right)$ and $\left(-3,0\right)$ is $9$.

Let $P \left(x, y\right)$ be the variable point.

Let $F_1 = \left(3,0\right)$ and $F_2 = \left(-3,0\right)$

Then as per question $\;$ $PF_1 + PF_2 = 9$

i.e. $\;$ $\sqrt{\left(x - 3\right)^2 + \left(y - 0\right)^2} + \sqrt{\left(x + 3\right)^2 + \left(y - 0\right)^2} = 9$

i.e. $\;$ $\sqrt{x^2 - 6x + 9 + y^2} - 9 = \sqrt{x^2 + 6x + 9 + y^2}$ $\;\;\; \cdots \; (1)$

Squaring both sides of equation $(1)$ gives

$x^2 - 6x + 9 + y^2 + 81 - 18 \sqrt{x^2 - 6x + 9 + y^2} = x^2 + 6 x + 9 + y^2$

i.e. $\;$ $- 12 x + 81 = 18 \sqrt{x^2 - 6x + 9 + y^2}$

i.e. $\;$ $-4 x + 27 = 6 \sqrt{x^2 - 6x + 9 + y^2}$ $\;\;\; \cdots \; (2)$

Squaring both sides of equation $(2)$ gives

$16x^2 + 729 - 216 x = 36 \left(x^2 - 6x + 9 + y^2\right)$

i.e. $\;$ $20 x^2 + 36 y^2 = 405$

i.e. $\;$ $\dfrac{x^2}{405 / 20} + \dfrac{y^2}{405 / 36} = 1$

i.e. $\;$ $\dfrac{x^2}{81 / 4} + \dfrac{y^2}{45 / 4} = 1$ $\;\;\; \cdots \; (3)$

Equation $(3)$ gives the required equation of locus.

Analytical Geometry - Conics - Ellipse

Find the equation of the ellipse if the foci are $\left(\pm 3, 0\right)$ and the length of the latus rectum is $\dfrac{32}{5}$

Let the foci be $\;$ $F_1 = \left(3,0\right)$; $\;\;$ $F_2 = \left(-3,0\right)$

Since the foci lie on the X axis, the major axis of the ellipse is along the X axis.

$\therefore$ $\;$ Let the equation of the required ellipse be

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

Now, length$\left(F_1 F_2\right) = 2ae$ $\;\;\; \cdots \; (2)$

length of latus rectum $= \dfrac{2b^2}{a}$ $\;\;\; \cdots \; (3)$

and $\;$ $b^2 = a^2 \left(1 - e^2\right)$ $\;\;\; \cdots \; (4)$

Given: $\;\;$ Length of latus rectum $= \dfrac{32}{5}$

$\implies$ $\dfrac{2b^2}{a} = \dfrac{32}{5}$ $\implies$ $b^2 = \dfrac{16a}{5}$ $\;\;\; \cdots \; (5a)$

Also, $\;$ $F_1 F_2 = \sqrt{\left(3 + 3\right)^2} = 6$

$\therefore$ $\;$ We have $\;\;$ $6 = 2ae$ $\implies$ $e = \dfrac{3}{a}$ $\;\;\; \cdots \; (5b)$

$\therefore$ $\;$ We have from equations $(4)$, $(5a)$ and $(5b)$,

$\dfrac{16a}{5} = a^2 \left(1 - \dfrac{9}{a^2}\right)$

i.e. $\;$ $\dfrac{16 a}{5} = a^2 - 9$

i.e. $\;$ $5a^2 - 16a - 45 = 0$

i.e. $\;$ $\left(a - 5\right) \left(5a + 9\right) = 0$

i.e. $\;$ $a = 5$ $\;\;\; \cdots \; (6)$ $\;$ or $\;$ $a = \dfrac{-9}{5}$ $\;$ (not an acceptable solution)

$\therefore$ $\;$ In view of equation $(6)$, we have from equation $(5a)$

$b^2 = \dfrac{16 \times 5}{5} = 16$

$\therefore$ $\;$ Substituting the values of $a$ and $b^2$ in equation $(1)$ gives the equation of the required ellipse as

$\dfrac{x^2}{5^2} + \dfrac{y^2}{16} = 1$

i.e. $\;$ $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$

Analytical Geometry - Conics - Ellipse

Find the equation of the ellipse if the center is $\left(3, -4\right)$, one of the foci is $\left(3 + \sqrt{3}, - 4\right)$ and $e = \dfrac{\sqrt{3}}{2}$

Let the center of the ellipse be $= C \left(h, k\right) = \left(3, -4\right)$

Let one of the foci be $= F_1 = \left(3 + \sqrt{3}, -4\right)$

Let the other focus be $= F_2 = \left(x_2, y_2\right)$

Now, center of the ellipse is the midpoint of $F_1 F_2$.

$\therefore$ $\;$ We have $\;$ $3 = \dfrac{3 + \sqrt{3} + x_2}{2}$; $\;$ $-4 = \dfrac{-4 + y_2}{2}$

i.e. $\;$ $x_2 = 3 - \sqrt{3}$; $\;$ $y_2 = -4$

$\therefore$ $\;$ $F_2 = \left(x_2, y_2\right) = \left(3 - \sqrt{3}, -4\right)$

$\therefore$ $\;$ From the given data, the major axis is parallel to the X axis.

$\therefore$ $\;$ Let the equation of the required ellipse be: $\;$ $\dfrac{\left(x - h\right)^2}{a^2} + \dfrac{\left(y - k\right)^2}{b^2} = 1$

Given: $\;$ $e = \dfrac{\sqrt{3}}{2}$

Now, $\;$ $F_1 F_2 = 2 a e$

i.e. $\;$ $\sqrt{\left(3 + \sqrt{3} - 3 + \sqrt{3}\right)^2 + \left(-4 + 4\right)^2} = 2 \times a \times \dfrac{\sqrt{3}}{2}$

i.e. $\;$ $2 \sqrt{3} = a \sqrt{3}$ $\implies$ $a = 2$

Further, $\;$ $b^2 = a^2 \left(1 - e^2\right)$

i.e. $\;$ $b^2 = \left(2\right)^2 \times \left[1 - \left(\dfrac{\sqrt{3}}{2}\right)^2\right]$

i.e. $\;$ $b^2 = 4 \times \dfrac{1}{4}$ $\implies$ $b^2 = 1$

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

$\dfrac{\left(x - 3\right)^2}{2^2} + \dfrac{\left(y + 4\right)^2}{2} = 1$

i.e. $\;$ $\dfrac{\left(x - 3\right)^2}{4} + \dfrac{\left(y + 4\right)^2}{1} = 1$

Analytical Geometry - Conics - Ellipse

Find the equation of the ellipse if the foci are $\left(\pm 3, 0\right)$ and the vertices are $\left(\pm 5, 0\right)$.

Let the foci be $F_1 = \left(3,0\right)$ and $F_2 = \left(-3,0\right)$

$\therefore$ $\;$ The foci $F_1$ and $F_2$ lie on the the X axis.

Let the equation of the ellipse be $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

Let the vertices be $V_1 = \left(5,0\right)$ and $V_2 = \left(-5,0\right)$

Now, $\text{length } \left(V_1 V_2\right) = 2a$

i.e. $\;$ $\sqrt{\left(5 + 5\right)^2} = 2a$

i.e. $\;$ $10 = 2a$ $\implies$ $a = 5$

and $\text{length } \left(F_1 F_2\right) = 2 a e$

i.e. $\;$ $\sqrt{\left(3 + 3\right)^2} = 2 \times 5 \times e$

i.e. $\;$ $6 = 10 e$ $\implies$ $e = \dfrac{3}{5}$

Also, $\;$ $b^2 = a^2 \left(1 - e^2\right)$

i.e. $\;$ $b^2 = \left(5\right)^2 \times \left[1 - \left(\dfrac{3}{5}\right)^2\right]$

i.e. $\;$ $b^2 = 25 \times \left[\dfrac{16}{25}\right]$ $\implies$ $b^2 = 16$

$\therefore$ $\;$ Required equation of ellipse is

$\dfrac{x^2}{5^2} + \dfrac{y^2}{16} = 1$

i.e. $\;$ $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$

Analytical Geometry - Conics - Ellipse

Find the equation of the ellipse if one of the foci is $\left(0,-1\right)$, the corresponding directrix is $\;$ $3x + 16 = 0$ $\;$ and $\;$ $e = \dfrac{3}{5}$

Let $\;$ $P \left(x,y\right)$ $\;$ be a moving point.

Focus $= F = \left(0,-1\right)$

Equation of directrix is $\;$ $3x + 16 = 0$

Draw $\;$ $PM$ $\;$ perpendicular to the directrix.

By definition, $\;$ $\dfrac{FP}{PM}= e$

$\therefore$ $\;$ $FP^2 = e^2 \; PM^2$

i.e. $\;$ $\left(x - 0\right)^2 + \left(y + 1\right)^2 = \left(\dfrac{3}{5}\right)^2 \times \left(\dfrac{\left|3x + 16\right|}{\sqrt{3^2}}\right)^2$

i.e. $\;$ $x^2 + y^2 + 2y + 1 = \dfrac{9}{25} \times \left(\dfrac{9x^2 + 96x + 256}{9}\right)$

i.e. $\;$ $25x^2 + 25y^2 + 50y + 25 = 9x^2 + 96x + 256$

i.e. $\;$ $16x^2 + 25y^2 - 96x + 50y - 231 = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is the equation of the required ellipse.

Analytical Geometry - Conics - Parabola

A cable of a suspension bridge is in the form of a parabola whose span is 40 m. The roadway is 5 m below the lowest point of the cable. If an extra support is provided across the cable 30 m above the ground level, find the length of the support if the height of the pillars are 55 m.

Take the lowest point on the suspension bridge as the vertex and let it be at the origin.

$\therefore$ $\;$ Vertex $= V = \left(0,0\right)$

Let $\;$ $RR'$ $\;$ be the roadway.

$PQ$ $\;$ is the span of the bridge $= 40$ m

Let extra support be provided at the points $AA'$.

$BA =$ height of point $A$ from the roadway $= 30$ m

$B'A' =$ height of point $A'$ from the roadway $= 30$ m

$PP'$, $\;$ $QQ' =$ pillars of height $55$ m

$\because$ $\;$ $PQ = 40$ m $\implies$ $VD' = 20$ m

$\therefore$ $\;$ Coordinates of $Q = \left(20, 50\right)$

Equation of the parabola is $\;\;$ $x^2 = 4ay$ $\;\;\; \cdots \; (1)$

$\because$ $\;$ Point $Q$ lies on equation $(1)$, we have

$\left(20\right)^2 = 4 \times a \times 50$ $\implies$ $a = 2$

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

$x^2 = 8y$ $\;\;\; \cdots \; (2)$

Let $VC' = \ell$ m.

From the figure, $\;$ $A'C' = 25$ m

Then the coordinates of point $A'$ are $\left(\ell, 25\right)$

$\because$ $\;$ Point $A'$ lies on the parabola given by equation $(2)$, we have

$\ell^2 = 8 \times 25 = 200$ $\implies$ $\ell = 10 \sqrt{2}$

$\therefore$ $\;$ From the figure, $AA' = 2 \ell = 20 \sqrt{2}$

$\therefore$ $\;$ Length of support $= 20 \sqrt{2}$ m

Analytical Geometry - Conics - Parabola

The focus of a parabolic mirror is at a distance of 8 cm from its center (vertex). If the mirror is 25 cm deep, find the diameter of the mirror.

Let the vertex of the parabolic mirror be at the origin i.e. $V \left(0,0\right)$

Let the equation of the parabolic mirror be: $\;\;$ $y^2 = 4ax$ $\;\;\; \cdots \; (1)$

Given: $\;$ Focus is at a distance of $8$ cm from the center.

$\therefore$ $\;$ $F = \left(a, 0\right) = \left(8, 0\right)$ $\;\;$ $\implies$ $a = 8$ $\;\;\; \cdots \; (2)$

Let the diameter of the parabolic mirror be $AB$.

Depth of the mirror $= 25$ cm

Let $p$ be the radius of the parabolic mirror.

Then, $\;$ $A = \left(25, p\right)$ $\;\;\; \cdots \; (3)$

Now $A$ lies on the parabola.

$\therefore$ $\;$ We have from equations $(1)$, $(2)$ and $(3)$,

$p^2 = 4 \times 8 \times 25 = 800$ $\;\;$ $\implies$ $p = 20 \sqrt{2}$

$\therefore$ $\;$ Diameter $AB = 2p = 40 \sqrt{2}$ cm

Analytical Geometry - Conics - Parabola

Find the axis, vertex, focus, equation of directrix, latus rectum and length of the latus rectum for the parabola $y^2 + 8x - 6y + 1 = 0$. Also sketch its graph.

Given equation of parabola is: $\;\;$ $y^2 + 8x - 6y + 1 = 0$

i.e. $y^2 - 6y = - 8x - 1$

i.e. $y^2 - 6y + 9 = - 8x - 1 + 9$

i.e. $\left(y - 3\right)^2 = - 8 x + 8$

i.e. $\left(y - 3\right)^2 = - 8 \left(x - 1\right)$ $\;\;\; \cdots \; (1)$

Let $X = x - 1$ $\;$ and $\;$ $Y = y - 3$

Then equation $(1)$ can be written as

$Y^2 = - 8 X$

i.e. $\left(Y - 0\right)^2 = - 4 \times 2 \times \left(X - 0\right)$ $\;\;\; \cdots \; (2)$

Comparing equation $(2)$ with the standard equation $\;\;$ $\left(Y - k\right)^2 = - 4 \times a \times \left(X - h\right)$ $\;$ gives

$\left(h, k\right) = \left(0, 0\right)$ $\;$ and $\;$ $a = 2$

$\therefore$ $\;$ For the given parabola,

	Referred to X, Y	Referred to x, y $x = X + 1$; $\;$ $y = Y + 3$
Axis of symmetry	X axis $\;$ $\left(Y = 0\right)$	$Y = 0 \implies y = 3$ i.e. $\;$ $y - 3 = 0$
Vertex	$V = \left(h, k\right) = \left(0,0\right)$	$X = 0 \implies x = 1$ $Y = 0 \implies y = 3$ $\therefore$ $\;$ $V = \left(1,3\right)$
Focus	$F = \left(-a, 0\right) = \left(-2, 0\right)$	$X = -2 \implies x = -1$ $Y = 0 \implies y = 3$ $\therefore$ $\;$ $F = \left(-1, 3\right)$
Equation of directrix	$X = a$ i.e. $\;$ $X = 2$	$X = 2 \implies x = 3$ i.e. $\;$ $x - 3 = 0$
Equation of latus rectum	$X = -a$ i.e. $\;$ $X = -2$	$X = -2 \implies x = -1$ i.e. $\;$ $x + 1 = 0$
Length of latus rectum	$4a = 4 \times 2 = 8$	$8$

Analytical Geometry - Conics - Parabola

Find the axis, vertex, focus, equation of directrix, latus rectum and length of the latus rectum for the parabola $\left(x - 4\right)^2 = 4 \left(y + 2\right)$. Also sketch its graph.

Given equation of parabola is: $\;\;$ $\left(x - 4\right)^2 = 4 \left(y + 2\right)$ $\;\;\; \cdots \; (1a)$

Let $\;$ $X = x - 4$ $\;$ and $\;$ $Y = y + 2$

Then equation $(1a)$ can be written as: $\;\;$ $X^2 = 4 Y$

i.e. $\left(X - 0\right)^2 = 4 \times 1 \times \left(Y - 0\right)$ $\;\;\; \cdots \; (1b)$

Comparing equation $(1b)$ with the standard equation: $\;\;$ $\left(X - h\right)^2 = 4 a \left(Y - k\right)$ $\;$ gives

$\left(h, k\right) = \left(0, 0\right)$; $\;\;$ $a = 1$

$\therefore$ $\;$ For the given parabola,

Analytical Geometry - Conics - Parabola

Find the axis, vertex, focus, equation of directrix, latus rectum and length of the latus rectum for the parabola $y^2 = -8 x$. Also sketch its graph.

Given equation of parabola is: $\;\;$ $y^2 = -8x$

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

Comparing equation $(1)$ with the standard equation: $\;$ $\left(y - k\right)^2 = - 4 a \left(x - h\right)$ $\;$ gives

$\left(h , k\right) = \left(0,0\right)$; $\;\;$ $a = 2$

$\therefore$ $\;$ For the given parabola,

Analytical Geometry - Conics - Parabola

Find the equation of the parabola if the vertex is $\left(3, -1\right)$, open rightward and the distance between latus rectum and the directrix is $4$.

The required parabola is open rightward.

$\therefore$ $\;$ Let its equation be

$\left(y - k\right)^2 = 4 a \left(x - h\right)$

Vertex is $= V \left(h, k\right) = \left(3, -1\right)$

Distance between latus rectum and directrix $= 2a$

Given: $\;\;$ $2a = 4 \implies a = 2$

$\therefore$ $\;$ Equation of parabola is

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

i.e. $\left(y + 1\right)^2 = 8 \left(x - 3\right)$

Analytical Geometry - Conics - Parabola

Find the equation of the parabola if the vertex is $\left(1,2\right)$ and latus rectum is $y = 5$.

Given: Vertex $= V = \left(h,k\right) = \left(1,2\right)$

Draw a perpendicular from V to the latus rectum.

It passes through the focus F.

$\therefore$ $\;$ F is $\left(1, 5\right)$

Also, $VF = a = 3$

From the given data, the parabola is open upwards.

$\therefore$ $\;$ The equation is of the form

$\left(x - h\right)^2 = 4a \left(y - k\right)$

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

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

i.e. $\left(x - 1\right)^2 = 12 \left(y - 2\right)$

Analytical Geometry - Conics - Parabola

Find the equation of the parabola if the vertex is $\left(0,0\right)$ and focus is $\left(0,-4\right)$.

From the given data, the parabola is open downwards.

Here, vertex $= V = \left(h,k\right) = \left(0,0\right)$

Focus $= F = \left(0, -4\right)$

Distance between vertex and focus $= VF = a$

i.e. $VF = a = \sqrt{\left(0 - 0\right)^2 + \left(0 + 4\right)^2} = 4$

The required equation of the parabola is of the form $\;\;$ $\left(x - h\right)^2 = - 4a \left(y - k\right)$

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

$\left(x - 0\right)^2 = - 4 \times 4 \times \left(y - 0\right)$

i.e. $x^2 = - 16 y$

Analytical Geometry - Conics - Parabola

Find the equation of the parabola if the focus is $\left(2, -3\right)$ and the directrix is $2y - 3 = 0$.

Let $P \left(x,y\right)$ be any point on the parabola with focus $F \left(2, -3\right)$ and directrix $\;$ $2y - 3 = 0$.

Draw $PM$ perpendicular to the directrix.

For a parabola, $\dfrac{FP}{PM} = \text{eccentricity} = e = 1$

$\implies$ $FP^2 = PM^2$ $\;\;\; \cdots \; (1)$

Now, $FP^2 = \left(x - 2\right)^2 + \left(y + 3\right)^2$ $\;\;\; \cdots \; (2a)$

and $PM^2 = \left(\pm \dfrac{2y - 3}{\sqrt{2^2}}\right) = \dfrac{\left(2y - 3\right)^2}{4}$ $\;\;\; \cdots \; (2b)$

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

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

i.e. $x^2 - 4x + 4 + y^2 + 6y + 9 = \dfrac{4y^2 - 12y + 9}{4}$

i.e. $4x^2 - 16 x + 4 y^2 + 24 y + 52 = 4y^2 - 12y + 9$

i.e. $4x^2 - 16x + 36 y + 43 = 0$ $\;\;\; \cdots \; (3)$

Equation $(3)$ is the equation of the required parabola.

Circle

Find the circle which cuts orthogonally each of the circles $\;$ $x^2 + y^2 + 2x + 4y + 1 = 0$, $\;$ $x^2 + y^2 - 4x + 3 = 0$ $\;$ and $\;$ $x^2 + y^2 + 6y + 5 = 0$.

Let the equation of the required circle be $\;$ $x^2 + y^2 + 2g_1 x + 2 f_1 y + c_1 = 0$ $\;\;\; \cdots \; (1)$

$\begin{aligned} \text{Equations of given circles are: } & x^2 + y^2 + 2x + 4y + 1 = 0 \;\;\; \cdots \; (2a) \\\\ & x^2 + y^2 - 4x + 3 = 0 \;\;\; \cdots \; (2b) \\\\ & x^2 + y^2 + 6y + 5 = 0 \;\;\; \cdots \; (2c) \end{aligned}$

Standard equation of circle: $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (3)$

Comparing equation $(2a)$ with equation $(3)$ gives

$2g_2 = 2 \implies g_2 = 1$; $\;\;$ $2f_2 = 4 \implies f_2 = 2$; $\;\;$ $c_2 = 1$ $\;\;\; \cdots \; (4a) $

Comparing equation $(2b)$ with equation $(3)$ gives

$2g_3 = -4 \implies g_3 = -2$; $\;\;$ $2f_3 = 0 \implies f_3 = 0$; $\;\;$ $c_3 = 3$ $\;\;\; \cdots \; (4b) $

Comparing equation $(2c)$ with equation $(3)$ gives

$2g_4 = 0 \implies g_4 = 0$; $\;\;$ $2f_4 = 6 \implies f_4 = 3$; $\;\;$ $c_4 = 5$ $\;\;\; \cdots \; (4c) $

Now, circles given by equations $(1)$ and $(2a)$ are orthogonal if $\;$ $2g_1 g_2 + 2 f_1 f_2 = c_1 + c_2$

i.e. $2 \times g_1 \times 1 + 2 \times f_1 \times 2 = c_1 + 1$

i.e. $2g_1 + 4f_1 = c_1 - 1$ $\;\;\; \cdots \; (5a)$

Circles given by equations $(1)$ and $(2b)$ are orthogonal if $\;$ $2g_1 g_3 + 2 f_1 f_3 = c_1 + c_3$

i.e. $2 \times g_1 \times \left(-2\right) + 2 \times f_1 \times 0 = c_1 + 3$

i.e. $-4g_1 = c_1 + 3$ $\implies$ $g_1 = \dfrac{- c_1 - 3}{4}$ $\;\;\; \cdots \; (5b)$

circles given by equations $(1)$ and $(2c)$ are orthogonal if $\;$ $2g_1 g_4 + 2 f_1 f_4 = c_1 + c_4$

i.e. $2 \times g_1 \times 0 + 2 \times f_1 \times 3 = c_1 + 5$

i.e. $6 f_1 = c_1 + 5$ $\implies$ $f_1 = \dfrac{c_1 + 5}{6}$ $\;\;\; \cdots \; (5c)$

$\therefore$ $\;$ In view of equations $(5b)$ and $(5c)$ equation $(5a)$ becomes

$2 \times \left(\dfrac{-c_1 - 3}{4}\right) + 4 \times \left(\dfrac{c_1 + 5}{6}\right) = c_1 + 1$

i.e. $\dfrac{-c_1 - 3}{2} + \dfrac{2 \left(c_1 + 5\right)}{3} = c_1 + 1$

i.e. $-3c_1 - 9 + 4 c_1 + 20 = 6 c_1 + 6$

i.e. $5 c_1 = 5$ $\implies$ $c_1 = 1$ $\;\;\; \cdots \; (6a)$

Substituting the value of $c_1$ in equation $(5b)$ gives

$g_1 = \dfrac{-1 - 3}{4} = -1$ $\;\;\; \cdots \; (6b)$

Substituting the value of $c_1$ in equation $(5c)$ gives

$f_1 = \dfrac{1 + 5}{6} = 1$ $\;\;\; \cdots \; (6c)$

$\therefore$ $\;$ In view of equations $(6a)$, $(6b)$ and $(6c)$, equation $(1)$ becomes

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

Equation $(7)$ gives the required circle orthogonal to the three given circles.

Circle

Show that the circles $\;$ $x^2 + y^2 - 2x + 6y + 6 = 0$ $\;$ and $\;$ $x^2 + y^2 - 5x + 6y + 15 = 0$ $\;$ touch each other.

$\begin{aligned} \text{Given equations of circles: } & x^2 + y^2 - 2x + 6y + 6 = 0 \;\;\; \cdots \; (1) \\\\ & x^2 + y^2 - 5x + 6y + 15 = 0 \;\;\; \cdots \; (2) \\\\ \text{Standard equation of circle: } & x^2 + y^2 + 2gx + 2fy + c = 0 \;\;\; \cdots \; (3) \end{aligned}$

Comparing equation $(1)$ with equation $(3)$ gives

$2 g_1 = -2 \implies g_1 = -1$; $\;\;$ $2f_1 = 6 \implies f_1 = 3$; $\;\;$ $c_1 = 6$

$\therefore$ $\;$ Center of equation $(1)$ is $= C_1 = \left(-g_1, -f_1\right) = \left(1, -3\right)$

Radius of equation $(1)$ is $= r_1 = \sqrt{g_1^2 + f_1^2 - c}$

i.e. $r_1 = \sqrt{\left(-1\right)^2 + \left(3\right)^2 - 6} = \sqrt{1 + 9 - 6} = 2$

Comparing equation $(2)$ with equation $(3)$ gives

$2g_2 = - 5 \implies g_2 = \dfrac{-5}{2}$; $\;\;$ $2 f_2 = 6 \implies f_2 = 3$; $\;\;$ $c_2 = 15$

$\therefore$ $\;$ Center of equation $(2)$ is $= C_2 = \left(-g_2, -f_2\right) = \left(\dfrac{5}{2}, -3\right)$

Radius of equation $(2)$ is $= r_2 = \sqrt{g_2^2 + f_2^2 - c_2}$

i.e. $r_2 = \sqrt{\left(\dfrac{-5}{2}\right)^2 + \left(3\right)^2 - 15} = \sqrt{\dfrac{25}{4} + 9 - 15} = \dfrac{1}{2}$

Now, $C_1 C_2 = \sqrt{\left(\dfrac{5}{2} - 1\right)^2 + \left(-3 + 3\right)^2} = \sqrt{\dfrac{9}{4}} = \dfrac{3}{2}$

and $r_1 - r_2 = 2 - \dfrac{1}{2} = \dfrac{3}{2}$

$\therefore$ $\;$ $C_1 C_2 = r_1 - r_2$

$\implies$ The circles given by equations $(1)$ and $(2)$ touch internally.

Circle

Find the value of $p$ so that the line $3x + 4y - p = 0$ is a tangent to $x^2 + y^2 - 64 = 0$.

Given equation of circle: $\;\;\;$ $x^2 + y^2 - 64 = 0$ $\;$ i.e. $\;$ $x^2 + y^2 = 64$ $\;\;\; \cdots \; (1)$

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

gives $\;\;\;$ $a^2 = 64$ $\;\;\; \cdots \; (2)$

Equation of given line: $\;\;\;$ $3x + 4y - p = 0$ $\;$ i.e. $\;$ $y = \dfrac{-3}{4}x + \dfrac{p}{4}$ $\;\;\; \cdots \; (3)$

Comparing equation $(3)$ with the standard equation of line $\;$ $y = mx + c$ $\;$ gives

$m = \dfrac{-3}{4}$ $\;\;\; \cdots \; (4a)$; $\;$ $c = \dfrac{p}{4}$ $\;\;\; \cdots \; (4b)$

Condition for a given line to be a tangent to a circle is: $\;$ $c^2 = a^2 \left(1 + m^2\right)$ $\;\;\; \cdots \; (5)$

$\therefore$ $\;$ In view of equations $(2)$, $(4a)$ and $(4b)$, equation $(5)$ becomes

$\dfrac{p^2}{16} = 64 \left[1 + \left(\dfrac{-3}{4}\right)^2\right]$

i.e. $\dfrac{p^2}{16} = 64 \left(1 + \dfrac{9}{16}\right)$

i.e. $\dfrac{p^2}{16} = 64 \times \dfrac{25}{16}$ $\implies$ $p = \pm 40$

Circle

Find the equation of the tangent lines to the circle $x^2 + y^2 = 9$ which are parallel to $2x + y - 3 = 0$.

Equation of given line: $\;\;\;$ $2x + y - 3 = 0$

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

$\therefore$ $\;$ Slope of given line $= m = -2$

Since the required tangents to the circle are parallel to the given line,

slope of required tangents $= m = -2$ $\;\;\; \cdots \; (2)$

Equation of given circle: $\;\;\;$ $x^2 + y^2 = 9$ $\;\;\; \cdots \; (3)$

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

we have $\;$ $a^2 = 9$ $\implies$ $a = \pm 3$

Equation of tangent (with slope m) to a circle $\;$ $x^2 + y^2 = a^2$ $\;$ is $\;$ $y = mx \pm a \sqrt{1 + m^2}$

$\therefore$ $\;$ Required equations of tangent lines are

$y = - 2x \pm 3 \sqrt{1 + \left(-2\right)^2}$

i.e. $y = -2x \pm 3 \sqrt{5}$

i.e. $2x + y = \pm 3 \sqrt{5}$

Circle

Find the coordinates of the point of intersection of the line $x + y =2$ with the circle $x^2 + y^2 = 4$.

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

Equation of given line: $\;\;\;$ $x + y = 2$ $\implies$ $y = 2 - x$ $\;\;\; \cdots \; (2)$

Substitute the value of $y$ from equation $(2)$ in equation $(1)$. We have

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

i.e. $x^2 - 4x = 0$

i.e. $x \left(x - 4\right) = 0$

i.e. $x = 0$ $\;$ or $\;$ $x = 4$

$\therefore$ $\;$ We have from equation $(2)$,

when $x = 0$, $y = 2 - 0 = 2$ and

when $x = 4$, $y = 2 - 4 = -2$

$\therefore$ $\;$ The required coordinates of the point of intersection of the line $x + y = 2$ with the circle $x^2 + y^2 = 4$ are $\left(0,2\right)$ and $\left(4,-2\right)$.

Circle

Find the equation of the tangent to the circle $\;$ $x^2 + y^2 - 4x + 8y -5 = 0$ $\;$ at $\;$ $\left(2,1\right)$.

Equation of given circle: $\;\;\;$ $x^2 + y^2 - 4x + 8y - 5 = 0$

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

$2g = -4$ $\implies$ $g = -2$; $\;\;\;$ $2f = 8$ $\implies$ $f = 4$; $\;\;\;$ $c = -5$

Equation of tangent to the circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ at the point $P \left(x_1, y_1\right)$ is

$xx_1 + yy_1 + g \left(x + x_1\right) + f \left(y + y_1\right) + c = 0$

Given: External point $P \left(x_1, y_1\right) = \left(2,1\right)$

$\therefore$ $\;$ Equation of required tangent is

$2x + y - 2 \left(x + 2\right)+ 4 \left(y + 1\right) - 5 = 0$

i.e. $2x + y - 2x - 4 + 4y + 4 - 5 = 0$

i.e. $5y - 5 = 0$

i.e. $y - 1 = 0$

Circle

Find the length of the tangent from the point $\left(1,2\right)$ to the circle $x^2 + y^2 -2x + 4y + 9 = 0$.

Comparing the given circle: $\;\;\;$ $x^2 + y^2 -2x + 4y + 9 = 0$

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

gives: $\;\;\;$ $2g = -2$ $\implies$ $g = -1$; $\;\;$ $2f = 4$ $\implies$ $f = 2$; $\;\;$ $c = 9$

Length of tangent from an external point $P \left(x_1, y_1\right)$ and touch the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ at point $T$ is

$PT = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$

$\therefore$ $\;$ Required length of tangent from $\left(1,2\right)$ to the given circle

$= \sqrt{1^2 + 2^2 - 2 \times 1 + 4 \times 2 + 9}$

$= \sqrt{1 + 4 - 2 + 8 + 9} = \sqrt{20} = 2 \sqrt{5}$ units

Circle

Find the

1. Cartesian equation of the circle whose parametric equations are $x = \dfrac{1}{4} \cos \theta$, $y = \dfrac{1}{4} \sin \theta$ and $0 < \theta \leq 2 \pi$.
2. parametric equation of the circle $4x^2 + 4y^2 = 9$.

1. Given: $\;\;\;$ $x = \dfrac{1}{4} \cos \theta$ $\implies$ $\cos \theta = 4x$ $\;\;\; \cdots \; (1)$
   
   and $y = \dfrac{1}{4} \sin \theta$ $\implies$ $\sin \theta = 4y$ $\;\;\; \cdots \; (2)$
   
   Squaring and adding equations $(1)$ and $(2)$ gives
   
   $\cos^2 \theta + \sin^2 \theta = 16 x^2 + 16 y^2$
   
   i.e. $16x^2 + 16 y^2 = 1$ $\;\;\; \cdots \; (3)$
   
   Equation $(3)$ is the required equation of the circle in Cartesian form.



2. Given equation of circle is: $\;\;\;$ $4x^2 + 4y^2 = 9$
   
   i.e. $x^2 + y^2 = \dfrac{9}{4}$
   
   i.e. $x^2 + y^2 = \left(\dfrac{3}{2}\right)^2$
   
   Comparing with the standard equation of circle $\;$ $x^2 + y^2 = a^2$ $\;$ gives the radius $= a = \dfrac{3}{2}$
   
   Parametric equations of the circle $\;$ $x^2 + y^2 = a^2$ $\;$ are $\;$ $x = a \cos \theta$; $y = a \sin \theta$
   
   $\therefore$ $\;$ The required parametric equations of the given circle are
   
   $x = \dfrac{3}{2} \cos \theta$; $y = \dfrac{3}{2} \sin \theta$

Circle

Find the equation of the circle with origin as center and passing through the vertices of an equilateral triangle whose median is of length $3a$.

In an equilateral triangle

1. the medians bisect the side which they meet at right angles.
2. the medians intersect at a point which is the circumcenter of the circumscribed circle.
3. the medians intersect at a point (centroid) which is $\dfrac{2}{3}^{rd}$ the length of the median, measured from the respective vertex.

Let $ABD$ be an equilateral triangle.

Let $DM$ be the median of the triangle.

Let $C$ be the centroid.

Then $C$ is the center of the circumcircle.

Given: $C = \left(0,0\right)$; $\;\;$ length of median $= DM = 3a$

$\therefore$ $\;$ $DC = \dfrac{2}{3} \times DM = \dfrac{2}{3} \times 3a = 2a$

But radius of required circle $= DC$

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

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

i.e. $x^2 + y^2 = 4a^2$

Circle

A circle of radius 2 lies in the first quadrant and touches both the axes of coordinates. Find the equation of the circle with center at $\left(6, 5\right)$ and touching the above circle externally.

Let the given circle (which is in the first quadrant), touch the X and Y axes at the points $A \left(a, 0\right)$ and $B \left(0, b\right)$ respectively.

Let C be the center of the given circle.

Then, $C = \left(a, b\right)$

Given: Radius of the circle $= 2$

i.e. $CA = CB = 2$

Now, $CA = \sqrt{\left(a - a\right)^2 + \left(b - 0\right)^2} = b$

and $\;$ $CB = \sqrt{\left(a - 0\right)^2 + \left(b - b\right)^2} = a$

$\therefore$ $\;$ $a = b = 2$

$\therefore$ $\;$ Center of the given circle $= C = \left(2, 2\right)$

Let the required circle touch the given circle at the point $P$.

Given: $\;$ center of the required circle $= C_1 = \left(6, 5\right)$

Radius of the given circle $= PC_1$

Now, $CC_1 = \sqrt{\left(6 - 2\right)^2 + \left(5 - 2\right)^2} = \sqrt{4^2 + 3^2} =5$

But $CC_1 = CP + PC_1$

$\therefore$ $\;$ $PC_1 = CC_1 - CP = 5 - 2 = 3$

$\therefore$ $\;$ Radius of the required circle $= 3$

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

$\left(x - 6\right)^2 + \left(y - 5\right)^2 = 3^2$

i.e. $x^2 + y^2 - 12 x - 10 y + 36 + 25 = 9$

i.e. $x^2 + y^2 - 12 x - 10 y + 52 = 0$

Circle

A circle has radius 3 and its center lies on the line $y = x - 1$. Find the equation of the circle if it passes through the point $\left(7,3\right)$.

Let the center of the circle be $= C = \left(h,k\right)$

The center lies on the line $\;\;\;$ $y = x - 1$

$\therefore$ $\;$ We have, $\;\;\;$ $k = h - 1$ $\;\;\; \cdots \; (1)$

Given: Radius of required circle $= 3$

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

$\left(x - h\right)^2 + \left(y - k\right)^2 = 3^2$

$\therefore$ $\;$ We have by equation $(1)$

$\left(x - h\right)^2 + \left(y - h + 1\right)^2 = 9$ $\;\;\; \cdots \; (2)$

The required circle passes through $\left(7,3\right)$.

$\therefore$ $\;$ We have from equation $(2)$,

$\left(7 - h\right)^2 + \left(3 - h + 1\right)^2 = 9$

i.e. $49 + h^2 - 14 h + 16 + h^2 - 8 h = 9$

i.e. $2h^2 - 22h + 56 = 0$

i.e. $h^2 - 11h + 28 = 0$

i.e. $\left(h - 7\right) \left(h - 4\right) = 0$

i.e. $h = 7$ $\;$ or $\;$ $h = 4$

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

when $h = 7$, $k = 6$ and

when $h = 4$, $k = 3$

$\therefore$ $\;$ The centers of the required circles are $\left(7, 6\right)$ and $\left(4, 3\right)$.

$\therefore$ $\;$ The equations of the required circles are:

when center is $\left(7, 6\right)$:

$\left(x - 6\right)^2 + \left(y - 7\right)^2 = 9$

i.e. $x^2 + y^2 - 12 x - 14 y + 36 + 49 = 9$

i.e. $x^2 + y^2 - 12 x - 14 y + 76 = 0$ $\;\;\; \cdots \; (3)$

when center is $\left(4, 3\right)$:

$\left(x - 4\right)^2 + \left(y - 3\right)^2 = 9$

i.e. $x^2 + y^2 - 8 x - 6 y + 16 + 9 = 9$

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

Equations $(3)$ and $(4)$ are the required equations of the circles.

Circle

Find the equation of the circle which passes through the origin and cuts-off intercepts of lengths 6 and 8 units from the positive parts of X and Y axes respectively.

Let the required circle cut the X axis at point A and the Y axis at point B.

Given: $\;\;\;$ X intercept $= 6$ units and Y intercept $= 8$ units

$\therefore$ $\;$ $A = \left(6,0\right)$ and $B = \left(0,8\right)$

$OA$ and $OB$ are the chords of the circle.

Let $C$ be the center of the circle.

Now, perpendicular from the center to a chord bisects the chord.

Draw perpendiculars from center $C$ to chords $OA$ and $OB$.

Then $M$ and $N$ are the midpoints of $OA$ and $OB$ respectively.

$\therefore$ $\;$ $M = \left(3,0\right)$ and $N = \left(0,4\right)$

$\therefore$ $\;$ Center $C = \left(3,4\right)$

Radius of the circle $= OC = \sqrt{\left(3 - 0\right)^2 + \left(4 - 0\right)^2} = 5$

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

$\left(x - 3\right)^2 + \left(y - 4\right)^2 = 5^2$

i.e. $x^2 + y^2 - 6x - 8y + 9 + 16 = 25$

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

Circle

Find the coordinates of the midpoint of the chord which the circle $x^2 + y^2 + 4x - 2y - 3 = 0$ cuts off on the line $y = x + 2$

Given equation of circle: $\hspace{1em}$ $x^2 + y^2 + 4x - 2y - 3 = 0$ $\;\;\; \cdots \; (1)$

Comparing with the standard equation: $\hspace{1em}$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$2g = 4 \implies g = 2$, $\;\;$ $2f = -2 \implies f = -1$, $\;\;$ $c = -3$

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

Equation of line is: $\hspace{1em}$ $y = x + 2$ $\;\;\; \cdots \; (2)$

Let the given line cut the circle at points $A$ and $B$.

Draw $CM$ perpendicular to chord $AB$.

Then $M$ is the midpoint of $AB$.

Slope of given line $= m = 1$

$\therefore$ $\;$ Slope of $CM = \dfrac{-1}{m}= -1$

$\therefore$ $\;$ Equation of CM is: $\hspace{1em}$ $\left(y - 1\right) = -1 \left(x + 2\right)$

i.e. $-y = x + 1$ $\;\;\; \cdots \; (3)$

Adding equations $(2)$ and $(3)$ gives

$0 = 2x + 3$ $\implies$ $x = \dfrac{-3}{2}$

$\therefore$ $\;$ We have from equation $(2)$, $\hspace{1em}$ $y = \dfrac{-3}{2} + 2 = \dfrac{1}{2}$

$\therefore$ $\;$ Coordinates of point $M$ are $\left(\dfrac{-3}{2}, \dfrac{1}{2}\right)$

Circle

Find the equation of the circle concentric with the circle $x^2 + y^2 - 4x - 6y - 3 = 0$ and which touches the Y axis.

Equation of given circle: $\hspace{1em}$ $x^2 + y^2 - 4x - 6y - 3 = 0$

Comparing with the standard equation of circle: $\hspace{1em}$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$2g = -4 \implies g = -2$, $\;\;$ $2f = -6 \implies f = -3$, $\;\;$ $c = -3$

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

$\because$ $\;$ The required circle is concentric with the given circle,

$\therefore$ $\;$ center of required circle $= C = \left(2,3\right)$

Let the required circle touch the Y axis at the point $A$.

Then, $A = \left(0,3\right)$

Radius of the required circle $= CA = \sqrt{\left(2 - 0\right)^2} = 2$

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

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

i.e. $x^2 + y^2 - 4x - 6y + 4 + 9 = 4$

i.e. $x^2 + y^2 - 4x - 6y + 9 = 0$

Circle

Find the equation of the circle which passes through two points on the X axis which are at distance 4 from the origin and whose radius is 5.

Let the required circle pass through points $A$ and $B$.

$A$ and $B$ are such that their distance from the origin is $4$ units.

$\therefore$ $\;$ Let $A = \left(-4,0\right)$ and $B = \left(4,0\right)$

From the figure, the center of the circle will lie on the Y axis.

Let the center of the circle be $C = \left(0,b\right)$.

Radius of the circle $= AC = BC = 5$ units (given)

Now, $AC = \sqrt{\left(0 + 4\right)^2 + \left(b - 0\right)^2} = 5$

i.e. $16 + b^2 = 25$ $\implies$ $b^2 = 9$ $\implies$ $b = \pm 3$

$\therefore$ $\;$ The centers of the required circles are $\left(0, \pm 3\right)$

$\therefore$ $\;$ Equations of the required circles are $\hspace{1em}$ $\left(x - 0\right)^2 + \left(y \mp 3\right)^2 = 5^2$

i.e. $x^2 + y^2 \mp 6y + 9 = 25$

i.e. $x^2 + y^2 \mp 6y - 16 = 0$

Circle

Find the equation of the circle which passes through the points $\left(4,2\right)$, $\left(-6,-2\right)$ and has its center on the Y axis.

The center of the circle lies on the Y axis.

Let center $= C = \left(0,b\right)$

The required circle passes through the points $A \left(4,2\right)$ and $B \left(-6,-2\right)$.

Radius of circle $= CA = CB$ $\;\;\; \cdots \; (1)$

Now, $CA = \sqrt{\left(4 - 0\right)^2 + \left(2 - b\right)^2} = \sqrt{20 - 4b + b^2}$ $\;\;\; \cdots \; (2a)$

and $CB = \sqrt{\left(0 + 6\right)^2 + \left(b + 2\right)^2} = \sqrt{40 + 4b + b^2}$ $\;\;\; \cdots \; (2b)$

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

$\sqrt{20 - 4b + b^2} = \sqrt{40 + 4b + b^2}$

i.e. $20 - 4b + b^2 = 40 + 4b + b^2$

i.e. $8b = -20$ $\implies$ $b = \dfrac{-5}{2}$

$\therefore$ $\;$ Center of required circle $= C = \left(0, \dfrac{-5}{2}\right)$

Substituting the value of $b$ in equation $(2a)$ gives

radius of required circle $= CA = \sqrt{20 - 4 \times \left(\dfrac{-5}{2}\right) + \left(\dfrac{-5}{2}\right)^2} = \sqrt{\dfrac{145}{4}}$

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

$\left(x - 0\right)^2 + \left(y + \dfrac{5}{2}\right)^2 = \dfrac{145}{4}$

i.e. $x^2 + y^2 + 5y + \dfrac{25}{4} = \dfrac{145}{4}$

i.e. $x^2 + y^2 + 5y - 30 = 0$

Circle

Find the equation of the circle which passes through the points $\left(4,1\right)$ and $\left(6,5\right)$ and has its center on the line $4x + y - 16 = 0$.

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

Equation $(1)$ passes through the point $\left(4,1\right)$.

$\therefore$ $\;$ We have $\;$ $\left(4\right)^2 + \left(1\right)^2 + 2 \times 4 \times g + 2 \times 1 \times f + c = 0$

i.e. $8g + 2f + 17 + c = 0$ $\;\;\; \cdots \; (2a)$

Equation $(1)$ also passes through the point $\left(6,5\right)$.

$\therefore$ $\;$ We have $\;$ $\left(6\right)^2 + \left(5\right)^2 + 2 \times 6 \times g + 2 \times 5 \times f + c = 0$

i.e. $12g + 10 f + 61 + c = 0$ $\;\;\; \cdots \; (2b)$

Solving equations $(2a)$ and $(2b)$ simultaneously we get,

$4g + 8f + 44 = 0$

i.e. $g + 2f + 11 = 0$ $\;\;\; \cdots \; (3a)$

Let the center of the required circle be $\left(-g, -f\right)$.

The center lies on $\;$ $4x + y - 16 = 0$

$\therefore$ $\;$ We have $\;$ $-4g - f - 16 = 0$ $\;\;\; \cdots \; (3b)$

Multiplying equation $(3b)$ with $2$ and adding to equation $(3a)$ gives

$-7g - 21 = 0$ $\implies$ $g = -3$

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

$-3 + 2f + 11 = 0$ $\implies$ $f = -4$

Substituting the values of $f$ and $g$ in equation $(2a)$ gives

$8 \times \left(-3\right) + 2 \times \left(-4\right) + 17 + c = 0$ $\implies$ $c = 15$

Substituting the values of $f$, $g$ and $c$ in equation $(1)$ gives the equation of the required circle as

$x^2 + y^2 - 6x - 8y + 15 = 0$

Circle

Find the equation of the circle concentric with the circle $\;$ $x^2 + y^2 - 6x + 12 y + 15 = 0$ $\;$ and of double its radius.

Given equation of circle: $\;\;\;$ $x^2 + y^2 - 6x + 12 y + 15 = 0$

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

$2g = -6 \implies g = -3$; $\;$ $2 f = 12 \implies f = 6$; $\;$ $c = 15$

$\therefore$ $\;$ Center of given circle $= C = \left(-g, -f\right) = \left(3, -6\right)$

Radius of given circle $= r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(-3\right)^2 + \left(6\right)^2 - 15} = \sqrt{30}$

$\because$ $\;$ The required circle is concentric with the given circle,

center of required circle $= C = \left(3, -6\right)$

Radius of required circle $= R = 2r = 2 \sqrt{30}$

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

$\left(x - 3\right)^2 + \left(y + 6\right)^2 = 120$

i.e. $x^2 + y^2 - 6x + 12 y + 9 + 36 - 120 = 0$

i.e. $x^2 + y^2 - 6x + 12y - 75 = 0$

Circle

Prove that the centers of the three circles $x^2 + y^2 = 1$, $\;$ $x^2 + y^2 + 6x - 2y = 1$ $\;$ and $\;$ $x^2 + y^2 - 12x + 4y = 1$ are collinear.

Equations of the circles are:

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

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

$x^2 + y^2 - 12x + 4y = 1$ $\;\;\; \cdots \; (3)$

Center of circle given by equation $(1)$ is $C_1 = \left(0,0\right)$

The standard equation of circle is: $\hspace{1em}$ $x^2 + y^2 + 2gx + 2 fy + c = 0$ $\;\;\; \cdots \; (4)$

Comparing equation of the circle $(2)$ with the standard equation $(4)$ gives

$2g = 6 \implies g = 3$; $\;\;\;$ $2f = -2 \implies f = -1$

$\therefore$ $\;$ Center of circle given by equation $(2)$ is $C_2 = \left(-g , -f\right) = \left(-3, 1\right)$

Comparing equation of the circle $(3)$ with the standard equation $(4)$ gives

$2g = -12 \implies g = -6$; $\;\;\;$ $2f = 4 \implies f = 2$

$\therefore$ $\;$ Center of circle given by equation $(3)$ is $C_3 = \left(-g , -f\right) = \left(6, -2\right)$

Now, slope of $C_1 C_2 = m_1 = \dfrac{1-0}{-3-0} = \dfrac{-1}{3}$

Slope of $C_2 C_3 = m_2 = \dfrac{- 2 - 1}{6 + 3} = \dfrac{-3}{9} = \dfrac{-1}{3}$

Slope of $C_3 C_1 = m_3 = \dfrac{-2 - 0}{6 - 0} = \dfrac{-2}{6} = \dfrac{-1}{3}$

$\because$ $\;$ $m_1 = m_2 = m_3$ $\implies$ $C_1$, $C_2$, $C_3$ are collinear.

Circle

If the equation $\;$ $3x^2 + \left(3 - p\right)xy + qy^2 - 2px = 8pq$ $\;$ represents a circle, find $p$ and $q$. Also determine the center and the radius of the circle.

The given equation is: $\hspace{1em}$ $3x^2 + \left(3 - p\right)xy + qy^2 - 2px = 8pq$ $\;\;\; \cdots \; (1)$

Standard equation of a circle is: $\hspace{1em}$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (2)$

Equation $(1)$ will represent a circle if the coefficient of the $xy$ term is $0$.

i.e. $3 - p = 0$ $\implies$ $p = 3$

Substituting the value of $p$ in equation $(1)$ gives $\;\;$ $3x^2 + qy^2 - 6x - 24q = 0$

i.e. $x^2 + \dfrac{q}{3} y^2 - 2 x - 8q = 0$ $\;\;\; \cdots \; (3)$

Since equations $(2)$ and $(3)$ both represent the equation of circle, comparing the coefficients of the $y^2$ term gives

$\dfrac{q}{3} = 1$ $\implies$ $q = 3$

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

$x^2 + y^2 - 2x -24 = 0$ $\;\;\; \cdots \; (4)$

Comparing equation $(4)$ with the standard equation of the circle [equation $(2)$] we get,

$2g = -2$ $\implies$ $g = - 1$, $\;\;\;$ $f = 0$, $\;\;\;$ $c = - 24$

Center of circle $= \left(-g, -f\right) = \left(1,0\right)$

Radius of circle $= \sqrt{g^2 + f^2 - c} = \sqrt{\left(-1\right)^2 + 0 + 24} = 5$ units

Circle

Find the equation of the circle with center $\left(2,3\right)$ and passing through the intersection of the lines $\;$ $3x - 2y -1 = 0$ and $4x + y - 27 = 0$

$\begin{aligned} \text{Equation of lines: } & 3x - 2y - 1 = 0 \;\;\; \cdots \; (1) \\\\ & 4x + y - 27 = 0 \;\;\; \cdots \; (2) \end{aligned}$

Multiplying equation $(2)$ by $2$ and adding to equation $(1)$ gives

$11x - 55 = 0$ $\implies$ $x = 5$

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

$15 - 2y -1 = 0$ $\implies$ $y = 7$

$\therefore$ $\;$ The point of intersection of lines $(1)$ and $(2)$ is $P \left(5,7\right)$

Given: Center of the circle $= C \left(2,3\right)$

Radius of the circle $= CP = \sqrt{\left(5 - 2\right)^2 + \left(7 - 3\right)^2} = 5$

$\therefore$ $\;$ Equation of circle with center $\left(2,3\right)$ and radius $5$ is

$\left(x - 2\right)^2 + \left(y - 3\right)^2 = 5^2$

i.e. $x^2 + y^2 - 4x - 6y - 12 = 0$

Circle

Find the equation of circles that touch both the axes and pass through the point $\left(-4,-2\right)$, in general form.

The required circle passes through the point $P \left(-4,-2\right)$.

$\because$ $\;$ Point $P$ is in the third quadrant, the required circle will touch the negative X and Y axes.

Let the required circle touch the X axis at the point $A \left(-a,0\right)$ and the Y axis at the point $B \left(0,-b\right)$.

Then, the center of the circle is $C \left(-a,-b\right)$.

Now, $AC = BC = PC =$ radius of the circle

$\implies$ $AC^2 = BC^2 = PC^2$ $\;\;\; \cdots \; (1)$

Now, $AC^2 = BC^2$ $\implies$ $b^2 = a^2$ $\implies$ $b = a$

$\therefore$ $\;$ Center of circle is $C \left(-a, -a\right)$

$\therefore$ $\;$ $PC^2 = AC^2$ $\implies$ $\left(-4 + a\right)^2 + \left(-2 + a\right)^2 = a^2$

i.e. $16 + a^2 - 8a + 4 + a^2 - 4a = a^2$

i.e. $a^2 - 12 a + 20 = 0$

i.e. $\left(a - 10\right) \left(a - 2\right) = 0$

i.e. $a = 10$ $\;$ or $\;$ $a = 2$

$\because$ $\;$ $b = a$, $b = 10$ $\;$ or $\;$ $b = 2$

$\therefore$ $\;$ Center of the circle is $C \left(-10,-10\right)$ or $C \left(-2,-2\right)$

and radius of circle $= a = 10$ or $a = 2$

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

for center $\left(-10,-10\right)$ and radius $=10$: $\;\;$ $\left(x + 10\right)^2 + \left(y + 10\right)^2 = 100$

i.e. $x^2 + y^2 + 20 x + 20 y + 100 = 0$

and for center $\left(-2,-2\right)$ and radius $=2$: $\;\;$ $\left(x + 2\right)^2 + \left(y + 2\right)^2 = 4$

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