Find the inverse point of (–2, 3) w.r.t the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-6\mathrm{y}+9=0$

Given circle is $\mathrm{s}={\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-6\mathrm{y}+9=0$
Given point is p(–2, 3)
$\text{ Polar of }\mathrm{p}(-2,3)\text{ w.r.t }\mathrm{s}=0\text{ is }{\mathrm{s}}_1=0$
$\begin{array}{l}\Rightarrow {\mathrm{xx}}_1+{\mathrm{yy}}_1-2\left(\mathrm{x}+{\mathrm{x}}_1\right)-3\left(\mathrm{y}+{\mathrm{y}}_1\right)+9=0\\ \Rightarrow -2\mathrm{x}+3\mathrm{y}-2(\mathrm{x}-2)-3(\mathrm{y}+3)+9=0\\ \Rightarrow -2\mathrm{x}+3\mathrm{y}-2\mathrm{x}+4-3\mathrm{y}-9+9=0\\ \Rightarrow -4\mathrm{x}+4=0\\ \Rightarrow \mathrm{x}-1=0\to 1\{\mathrm{ax}+\mathrm{by}+\mathrm{c}=0\text{ from }\}\end{array}$
Let Q(h, k) be the inverse point of p w.r.t s = 0
Them, Q = foot of the
ar form p(–2, 3) on polar (1)
$\begin{array}{l}\frac{\mathrm{h}-{\mathrm{x}}_1}{\mathrm{a}}=\frac{\mathrm{k}-{\mathrm{y}}_1}{\mathrm{b}}=\frac{-\left({\mathrm{ax}}_1+{\mathrm{by}}_1+\mathrm{c}\right)}{{\mathrm{a}}^2+{\mathrm{b}}^2}\\ \Rightarrow \frac{\mathrm{h}+2}{1}=\frac{\mathrm{k}-3}{0}=\frac{-(-2-1)}{1^2+0^2}\\ \Rightarrow \frac{\mathrm{h}+2}{1}=3 \begin{array}{c}\frac{\mathrm{k}-3}{0}=3\\ \mathrm{k}-3=0\\ \Rightarrow \mathrm{k}=3\end{array}\\ \Rightarrow \mathrm{h}=1\end{array}\begin{array}{c} \\ \\ \end{array}$
$\therefore \text{ Inverse point of }\mathrm{p}\text{ is }\mathrm{Q}(\mathrm{h},\mathrm{k})=\mathrm{Q}(1,3)$