Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ…

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
  1. $\sqrt{\mathrm{PQ} \cdot \mathrm{RS}}$
  2. $\frac{\mathrm{PQ}+\mathrm{RS}}{2}$
  3. $\frac{2 \cdot \mathrm{PQ} \cdot \mathrm{RS}}{\mathrm{PQ}+\mathrm{RS}}$
  4. $\sqrt{\frac{\mathrm{PQ}^2+\mathrm{RS}^2}{2}}$

Solution


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 $\begin{aligned} & \frac{\mathrm{RS}}{2 r}=\frac{2 r}{\mathrm{PQ}} \\ \therefore \quad & 2 \mathrm{r}=\sqrt{\mathrm{PQ} \cdot \mathrm{RS}} \end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 2)

Practice more Circle questions on Aicharya