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Q.
The inverse point of $(1,2)$ with respect to the circle $x^{2}+y^{2}-4 x-6 y+9=0$ is
EAMCETEAMCET 2007
Solution:
Inverse point of $P(1,2)$ w.r.t. the circle is the foot of the perpendicular of $P$ on the polar of $P$.
Given circle is $x^{2}+y^{2}-4 x-6 y+9=0$ Polar of $P(1,2)$ is
$x \cdot 1+y \cdot 2-2(x+1)-3(y+2)+ 9=0$
$\Rightarrow x+2 y-2 x-2-3 y-6+9=0$
$\Rightarrow x+y-1=0$
Verifying inverse point of $P$ is $(0,1)$.