Coordinate Geometry - Circle

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

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

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

Radius of the circle $= a = 7$

Let the equation of the required circle be

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

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

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

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

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

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

Coordinate Geometry - Circle

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

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

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

Then equation $(1)$ should be identical with

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

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

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

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

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

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

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

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

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

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

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

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

$\implies$ $R = 25$

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

$625 - a^2 = 141$

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

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

Coordinate Geometry - Circle

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

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

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

Then equation $(1)$ should be identical with

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

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

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

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

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

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

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

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

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

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

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

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

$\implies$ $R = 4$

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

$16 - a^2 = 7$

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

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

Coordinate Geometry - Circle

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

Given equation of system of circles

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

Equation $(1)$ can be written as

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Coordinate Geometry - Circle

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

The given circles are

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\therefore \;$ The required tangents are

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

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

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

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

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

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

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

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

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

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

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

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

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

$\therefore \;$ The required tangents are

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

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

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

Coordinate Geometry - Circle

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

The given circles are

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

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

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

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

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

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

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

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

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

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

$\therefore \;$ We have,

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

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

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

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

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

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

Coordinate Geometry - Circle

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

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

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

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

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

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

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

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

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

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

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

$\therefore \;$ Equation of required circle is

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

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

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

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

Coordinate Geometry - Circle

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

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

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

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

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

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

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

$\therefore \;$ We have

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

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

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

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

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

which is the required equation of circle.

Coordinate Geometry - Circle

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

Given circles:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\therefore \;$ Required equation of circle is

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

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

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

Coordinate Geometry - Circle

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

Given circles:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Hence proved.

Coordinate Geometry - Circle

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

Given circles:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Coordinate Geometry - Circle

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

Equation of system of circles is

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

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

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

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

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

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

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

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

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

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

Coordinate Geometry - Circle

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

The extremities of the diameter are

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

$\therefore \;$ Equation of required circle is

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

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

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

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

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

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

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

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

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

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

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

Equations of required tangents are

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

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

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

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

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

Coordinate Geometry - Circle

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

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

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

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

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

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

Slope of the given line $= m_1 = 2$

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

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

Equations of tangents are

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

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

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

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

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

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

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

Coordinate Geometry - Circle

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

Given circles:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\therefore \;$ Required equation of circle is

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

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

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

Coordinate Geometry - Circle

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Coordinate Geometry - Circle

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Coordinate Geometry - Circle

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

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

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

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

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

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

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

Area of $\triangle OQR$

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

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

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

Coordinate Geometry - Circle

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Hence proved.

Coordinate Geometry - Circle

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

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

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

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

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

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

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

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

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

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

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

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

Solving, we get,

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

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

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

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

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

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

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