Find diameter PR of circle given tangents PQ and RS with their intersection X on circumference

Question

Let $PQ$ and $RS$ be tangents at the extremities of the diameter $PR$ of a circle of radius $r$. If $PS$ and $RQ$ intersect at a point $X$ on the circumference of the circle then find $2r$ in terms of $PQ, RS$

My attempt is as follows:-

In right angle $\triangle PXR$

$$PX^2+RX^2=PR^2$$ $$4r^2=PX^2+RX^2\tag{1}$$

$$\triangle PXQ \sim \triangle SXR$$, hence

$$\dfrac{SX}{PX}=\dfrac{RX}{QX}=\dfrac{SR}{PQ}$$

$$\dfrac{PS}{PX}=\dfrac{SR+PQ}{PQ}$$ $$PX=\dfrac{PS\cdot PQ}{SR+PQ}\tag{2}$$

In the same way

$$RX=\dfrac{RQ\cdot RS}{SR+PQ}\tag{3}$$

Placing all this in equation $(1)$

$$4r^2(PQ+RS)^2=PS^2\cdot PQ^2+RQ^2\cdot RS^2\tag{4}$$

Now we have to eliminate $PS$ and $RQ$

$$PS^2=PR^2+RS^2\tag{5}$$ $$RQ^2=PQ^2+PR^2\tag{6}$$

Placing all this in equation $(4)$

$$4r^2(PQ+RS)^2=(PR^2+RS^2)PQ^2+(PR^2+PQ^2)RS^2$$ $$4r^2(PQ^2+RS^2+2PQ\cdot RS-PQ^2-RS^2)=2PQ^2RS^2$$ $$2r=\sqrt{PQ\cdot RS}$$

Now one can see that the solution got very long + it took me some time to think, am I missing some property here which can come handy?

Answer 1

It'd much be simpler if you could identify the relevant similar triangles.

Because of the shared arc PX and PQ || RS, we have ∠PRQ = ∠QPX = ∠PSR. Then, the right triangles PQR and PSR are simpler, which leads to

$$\frac{PR}{PQ}=\frac{SR}{PR}$$

and the result follows.

Answer 2

Let $\angle RPS=\theta.$ Then $\angle PRS=90^\circ-\theta$

In $\triangle RPS$ and $\triangle RPQ$

$\displaystyle \tan \theta = \frac{RS}{2r}$ and $\displaystyle \tan (90^\circ-\theta)=\cot \theta =\frac{PQ}{2r}$

So we have $$\tan \theta \cdot \cot \theta=1 =\frac{RS\cdot PQ}{(2r)^2}\Rightarrow 2r=\sqrt{PQ\cdot RS.}$$