Finding circumcentre Tangents are draw from $P(2,3)$ to $x^2+y^2=4$ meeting at $Q,R$ on circle. Parallelogram $PQSR$ is completed. Find the circumcentre of triangle $QSR$.

My attempt: Clearly, the parallelogram is a rhombus. $S$ will be mirror image of $P$ in chord of contact. But, I want to know a good way of finding the circumcentre other than it lies on $PS$ and dropping perpendiculars and other boring stuff. Maybe a nice geometrical solution? Or combining it with coordinate?

Here's an overview ...
Note that the circumcenter (call it $C$) of $\triangle PQR$ is easy to find: It's the intersection of $\overline{PS}$ (a line through the origin) and the (horizontal) perpendicular bisector of (vertical) segment $\overline{PQ}$ with known endpoints. Now, the circumcenter you want (call it $D$) is clearly the reflection of $C$ in $\overline{QR}$; this implies that $\overline{CQ} \cong \overline{DQ}$. Thus, the circle with center $Q$ through $C$ meets $\overline{PS}$ at $D$.

Details ...
$$\overline{PS}:\;\; y = \frac{3}{2}x,\qquad\text{perp bis of } \overline{PQ}:\;\; y = \frac{3}{2}\qquad \implies \qquad C = \left( 1, \frac{3}{2} \right)$$
Then, writing $D = \left(d, \frac{3}{2}d\right)$, we have
$$\begin{align}
|\overline{DQ}|^2 = |\overline{CQ}|^2 \quad &\implies \quad\left(d-2\right)^2 + \left(\frac{3}{2}d - 0\right)^2 = \left( 1 - 2 \right)^2 + \left( \frac{3}{2} - 0 \right)^2 \\[6pt]
&\implies 13 d^2 - 16 d + 3 = 0 \\[6pt]
&\implies d = \frac{3}{13} \qquad (\text{or } d = 1, \text{ but that gives } C ) \\[6pt]
&\implies D = \left(\frac{3}{13}, \frac{9}{26}\right) \qquad \square
\end{align}$$
