If the inverse point of the point $(-1,1)$ with respect to the circle $x^2+y^2-2 x+2 y-1=0$ is $(p, q)$,…
If the inverse point of the point $(-1,1)$ with respect to the circle $x^2+y^2-2 x+2 y-1=0$ is $(p, q)$, then $p^2+q^2=$
$\frac{1}{16}$
$\frac{1}{8}$
$\frac{1}{4}$
$\frac{1}{2}$
Solution
The equation of pole w.r.t. point $(-1,1)$ to the circle $x^2+y^2-2 x+2 y-1=0$ is
$\begin{aligned}
& -x+y-(x-1)+(y+1)-1=0 \\
& \Rightarrow 2 x-2 y=1
\end{aligned}$
Since, inverse point of $(-1,1)$ is foot of the perpendicular from $(-1,1)$ to the line $2 x-2 y-1=0$
$\begin{gathered}
\therefore(p, q)=\left(\frac{b^2 x_1+a b y_1-a c}{\left(a^2+b^2\right)}, \frac{b^2 y_1-a b x_1-b c}{\left(a^2+b^2\right)}\right) \\
=\left(\frac{4(-1)+(2)(2)(1)-(2)(-1)}{8}, \frac{4(1)-(2)(-2)(-1)-(-2)(-1)}{8}\right) \\
\quad(p, q)=\left(\frac{1}{4}, \frac{-1}{4}\right) \Rightarrow p^2+q^2=\frac{1}{16}+\frac{1}{16}=\frac{1}{8}
\end{gathered}$