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Q.
The equation of the polar of the point $(2,-1)$ with respect to the circle $x^{2}+y^{2}-3 x+4 y-8=0$, is
Conic Sections
Solution:
Given circle is $x^{2}+y^{2}-3 x+4 y-8=0$...(i)
Given point is $(2,-1)$ let $P =(2,-1)$. Now equation of the
polar of point $P$ with respect to circle (i)
$x .2+y(-1)-3\left(\frac{x+2}{2}\right)+4\left(\frac{y-1}{2}\right)-8=0$
or $4 x-2 y-3 x-6+4 y-4-16=0$
or $x+2 y -26=0$