Find the pole of the straight line $\;$ $3x + 4y - 12 = 0$ $\;$ with respect to the circle $\;$ $x^2 + y^2 = 24$.
Given line: $\;$ $3x + 4y - 12 = 0$ $\;\;\; \cdots \; (1)$
Given circle: $\;$ $x^2 + y^2 = 24$ $\;\;\; \cdots \; (2)$
Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 = r^2$
gives $\;$ $r^2 = 24$
Let $\;$ $\left(h, k\right)$ $\;$ be the pole of equation $(1)$ with respect to equation $(2)$.
Then equation $(1)$ must be identical with the polar of $\;$ $\left(h, k\right)$ $\;$ with respect to equation $(2)$
i.e. $\;$ with $\;$ $xh + yk = r^2$
i.e. $\;$ with $\;$ $xh + yk = 24$
i.e. $\;$ with $\;$ $xh + yk - 24 = 0$ $\;\;\; \cdots \; (3)$
$\therefore \;$ Comparing equations $(1)$ and $(3)$ gives
$\dfrac{h}{3} = \dfrac{k}{4} = \dfrac{-24}{-12}$
Solving, we get,
$h = 6, \; k = 8$
$\therefore \;$ The pole of equation $(1)$ with respect to equation $(2)$ is $\left(6, 8\right)$.