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 2 r equals

Consider the given figure,
In \(\triangle \mathrm{PXR}, \angle \mathrm{X}=90^{\circ}\)
\(\therefore \quad \angle \mathrm{PRX}=90^{\circ}-\theta\)...(i)
\(\therefore \quad\) In \(\triangle\) PRS,
\(\begin{array}{ll}
& \tan \theta=\frac{\mathrm{RS}}{\mathrm{PR}}=\frac{\mathrm{RS}}{2 \mathrm{r}} ...(ii)\\
& \text { In } \triangle \mathrm{PRQ}, \angle \mathrm{PRQ}=90^{\circ}-\theta \\
\therefore \quad & \tan \left(90^{\circ}-\theta\right)=\frac{\mathrm{PQ}}{\mathrm{PR}}=\frac{\mathrm{PQ}}{2 \mathrm{r}} \\
\therefore \quad & \cot \theta=\frac{\mathrm{PQ}}{2 \mathrm{r}} \\
\therefore \quad & \tan \theta=\frac{2 \mathrm{r}}{\mathrm{PQ}}...(iii)
\end{array}\)
\(\therefore \quad\) from (ii) and (iii), we get
\(\begin{aligned}
& \frac{\mathrm{RS}}{2 r}=\frac{2 r}{\mathrm{PQ}} \\
\therefore \quad & 2 \mathrm{r}=\sqrt{\mathrm{PQ} \cdot \mathrm{RS}}
\end{aligned}\)