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)

Answer 2

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