Question

The inverse point of $(1,2)$ with respect to the circle
$x^2+y^2-4x-6y+9=0$ is

• A. (0,0)
• B. (1,0)
• C. (0,1)
• D. (1,1)

Answer 1

Answer: (C)

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 $x^2+y^2-4x-6y+9=0$ Polar of $P(1,2)$ is
$x\cdot 1+y\cdot 2-2(x+1)-3(y+2)+9=0$
$\Rightarrow x+2y-2x-2-3y-6+9=0$
$\Rightarrow x+y-1=0$
Verifying inverse point of $P$ is $(0,1)$.