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?
 A: 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
$$\begin{align}
|\overline{DQ}|^2 = |\overline{CQ}|^2 \quad &\implies \quad\left(d-2\right)^2 + \left(\frac{3}{2}d - 0\right)^2 = \left( 1 - 2 \right)^2 + \left( \frac{3}{2} - 0 \right)^2 \\[6pt]
&\implies 13 d^2 - 16 d + 3 = 0 \\[6pt]
&\implies d = \frac{3}{13} \qquad (\text{or } d = 1, \text{ but that gives } C ) \\[6pt]
&\implies D = \left(\frac{3}{13}, \frac{9}{26}\right) \qquad \square
\end{align}$$
A: 
 This is meant to be a useful drawing for the problem, not an answer. You should be able to take it off the screen to use for a proof. 
A: The equation of the circle with center $(0, 0)$ and radius $r$ is 
$$x^2 + y^2 = r^2.$$
The equation of tangent to this circle at the point $(x_p, y_p)$ on the circle is 
$$x_px + y_py = r^2,$$
then the equation of tangent at the point $(2,3)$ is
$$2x + 3y = 4\quad\Rightarrow\quad y=\frac{4-2x}3.$$
Now, we determine the points of tangency (coordinates of $R$ and $Q$).
$$
\begin{align}
x^2+y^2&=4\\
x^2+\left(\frac{4-2x}3\right)^2&=4\\
x_1=2\quad&;\quad x_2=-\frac{10}{13}\\
&\text{and}\\
y_2=0\quad&;\quad y_2=\frac{24}{13}.\\
\end{align}
$$
We obtain $Q(2,0)$ and $R\left(-\dfrac{10}{13},\dfrac{24}{13}\right).$ Since $PQRS$ is a parallelogram, then $PQ\parallel RS$ and $PR\parallel QS$. It's easy to notice that $PQ$ is perpendicular with $x$-axis, therefore $RS$ is also perpendicular with $x$-axis and line $RS: x=-\dfrac{10}{13}$.
Line $PQ:\ x=2$
Line $PR:$
$$
\begin{align}
\frac{y-y_P}{y_R-y_P}&=\frac{x-x_P}{x_R-x_P}\\
\frac{y-3}{\frac{24}{13}-3}&=\frac{x-2}{-\frac{10}{13}-2}\\
y&=\frac{5}{12}x+\frac{13}{6}
\end{align}
$$
Line $RQ:y=-\dfrac{2}{3}x+\dfrac{4}{3}$
Since $PR\parallel QS$, then gradient line $PR$ is equal to gradient line $QS$, $m=\dfrac{5}{12}$. Therefore, line $QS:$
$$
\begin{align}
y-y_Q&=m(x-x_Q)\\
y-0&=\dfrac{5}{12}(x-2)\\
y&=\dfrac{5}{12}x-\dfrac{5}{6}\\
\end{align}
$$
The coordinate of $S$ is the intersection point line $RS$ and $QS$. Hence $x_S=-\dfrac{10}{13}$ and
$$
y_S=\dfrac{5}{12}x_S-\dfrac{5}{6}=-\dfrac{15}{13}.
$$
Now, we determine line that is perpendicular to line $RQ$ and passes through point $S$, let's call this line as $SS'$. The gradient of $SS'$ is $-\dfrac1{m_{RQ}}=\dfrac32$, then the equation is
$$
\begin{align}
y-y_S&=m(x-x_S)\\
y+\dfrac{15}{13}&=\dfrac32\left(x+\dfrac{10}{13}\right)\\
y&=\frac32 x.
\end{align}
$$
The midle point of line $RS$ is
$$
y_M=\frac12(y_R+y_S)=\frac12\left(\dfrac{24}{13}-\dfrac{15}{13}\right)=\frac9{26}.
$$
It is clearly that the circumcentre $\Delta QSR$ on the intersection point of line $SS'$ and line $QM$. I think you can handle it the rest.
P.S. : Unfortunately, I cannot find the fastest way than this.
