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)

$(0,1)$

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, $\frac{\alpha -1}{1}=\frac{\beta -2}{1}=-\frac{1.1+1.2-1}{1^2+1^2}$
$\Rightarrow \alpha -1=\beta -2=-1$
$\Rightarrow \alpha =0,\beta =1$
Hence, required point is $(0,1)$.