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 $\left(\dfrac{-11}{7}, \dfrac{4}{7}\right)$.