We have, $\angle A P B=\frac{\pi}{4}$
$\begin{array}{ll}\Rightarrow & \angle A O B=2 \angle A P B=2 \times \frac{\pi}{4}=\frac{\pi}{2} \\ \therefore & \angle O A B=\angle O B A=\frac{\pi}{4} \\ \therefore & O A=O B \Rightarrow O A^2=O B^2 \\ \Rightarrow & (h+1)^2+(k-3)^2=(h-5)^2+(k-3)^2 \\ \Rightarrow & (h+1)^2=(h-5)^2 \\ \Rightarrow & h^2+2 h+1=h^2-10 h+25 \\ \Rightarrow & 12 h=24 \Rightarrow h=2\end{array}$
Since, $\angle A O B=\frac{\pi}{2}$
$\therefore \quad O A \perp O B$
$\Rightarrow$ Slope of $O A \times$ Slope of $O B=-1$
$\begin{array}{ll}\Rightarrow & \frac{k-3}{h+1} \times \frac{k-3}{h-5}=-1 \\ \Rightarrow \quad(k-3)^2=-(h+1)(h-5) \\ \Rightarrow \quad(k-3)^2=-(2+1)(2-5) \\ \Rightarrow \quad(k-3)^2=9 \Rightarrow k-3= \pm 3 \Rightarrow k=3 \pm 3 \\ \Rightarrow \quad k=0,6\end{array}$
$\therefore$ Centre of circle is $(2,0)$ or $(2,6)$
When centre is $(2,0)$
$\therefore$ Radius of circle $=O A$
$=\sqrt{(2+1)^2+(0-3)^2}=\sqrt{9+9}=\sqrt{18}$
When centre is $(2,6)$
$\begin{aligned} \therefore \text { Radius of circle } & =\sqrt{(2+1)^2+(6-3)^2} \\ & =\sqrt{9+9}=3 \sqrt{2}\end{aligned}$
$\therefore$ Equation of circle is
$\begin{aligned} & (x-2)^2+(y-0)^2=18 \\ & \Rightarrow \quad x^2+y^2-4 x-22=0 \\ & \text { or } \quad(x-2)^2+(y-6)^2=18 \\ & \Rightarrow \quad x^2+y^2-4 x-12 y+22=0 \\ & \end{aligned}$