Question

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

• A. $1,\frac{1}{2}$
• B. (2, 1)
• C. (0, 1)
• D. (1, 0)

Answer 1

Answer: (C)

The equation of pole with respect to the point $(1,2)$ to the circle
$x^2+y^2-4x-6y+9=0$ is
$x+2y-2(x+1)-3(y+2)+9=0$
$\Rightarrow x+y-1=0$
Since, the inverse of the point $(1,2)$ is the foot $(\alpha ,\beta )$ of the perpendicular from the point $(1,2)$ to the line $x+y-1$
Therefore, $\frac{\alpha -1}{1}=\frac{\beta -2}{1}=-\frac{1.1+1.2-1}{1^2+1^2}$
$\Rightarrow \alpha -1=\beta -2=-1$
$\Rightarrow \alpha =0,\beta =1$