Let $P$ and $Q$ be two distinct points on a circle which has center at $C\left(2,3\right)$ and which passes through origin $O$. If $OC$ is perpendicular to both the line segments $CP$ and $CQ$, then the set $\left\{P,Q\right\}$ is equal to

Slope of $\mathrm{OC}=\frac{3}{2}$

Since the line $OC$ is perpendicular to the line $PQ$ then the slope of the line $PQ$ is $-\frac{2}{3}$  

               $\mathrm{tan\theta }=-\frac{2}{3}\Rightarrow \sin \theta =\frac{2}{\sqrt{13}},\cos \theta =-\frac{3}{\sqrt{13}}$

Using symmetric from the line, 

The coordinates of point $P$ is

             $\begin{array}{rcl}P& =& \left(2+\sqrt{13}\cos \theta ,3+\sqrt{13}\sin \theta \right)\\ & =& \left(2+\sqrt{13}\left(-\frac{3}{\sqrt{13}}\right),3+\sqrt{13}\left(\frac{2}{\sqrt{13}}\right)\right)\\ & =& \left(2-3,3+2\right)\\ & =& \boxed{\left(-1,5\right)}\end{array}$

and the other point is Q

             $\begin{array}{rcl}Q& =& \left(2-\sqrt{13}\cos \theta ,3-\sqrt{13}\sin \theta \right)\\ & =& \left(2-\sqrt{13}\left(-\frac{3}{\sqrt{13}}\right),3-\sqrt{13}\left(\frac{2}{\sqrt{13}}\right)\right)\\ & =& \left(2+3,3-2\right)\\ & =& \boxed{\left(5,1\right)}\end{array}$