Given the circle has centre $\left(2,3\right)$ and it passes through the point $\left(0,0\right).$

So, the radius of the circle is $\sqrt{2^2+3^2}=\sqrt{13}$

The equation of the line which passes through the point $\left(0,0\right) \& \left(2,3\right)$ is $y-0=\frac{3-0}{2-0}\left(x-0\right)\Rightarrow y=\frac{3}{2}x$

So, the slope of $OC=\frac{3}{2}$

From the diagram, the line $PQ$ is perpendicular to the line $OC.$

So, the slope of $PQ$ is $-\frac{2}{3}$

So, $\tan \theta =-\frac{2}{3}$

Using symmetric from of line

The coordinate of $P=\left(2+\sqrt{13}\cos \theta ,3+\sqrt{13}\sin \theta \right)$

$=\left(2+\sqrt{13}·\left(-\frac{3}{\sqrt{13}}\right),3+\sqrt{13}\left(\frac{2}{\sqrt{13}}\right)\right)$

$=\left(-1,5\right)$

The coordinate of $\mathrm{Q}=\left(2-\sqrt{13}\cos \theta ,3-\sqrt{13}\sin \theta \right)$

$=\left(2-\sqrt{13}·\left(-\frac{3}{\sqrt{13}}\right),3-\sqrt{13}\left(\frac{2}{\sqrt{13}}\right)\right)$

$=\left(5,1\right)$