Let $P Q$ and $R S$ be tangents at the extremities of a diameter $P R$ of a circle of radius $r$ such that $P S$ and $R Q$ intersect at a point $X$ on the circumference of the circle, then $2 r$ equals

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Aicharya · Accepted Answer

According to the question, from the diagram, in $\triangle P Q R$ $\tan \theta=\frac{P Q}{P R} \Rightarrow P R=P Q \cot \theta$...(i)

and in $\triangle P R S$, $\begin{aligned} & \tan \left(90^{\circ}-\theta\right) =\frac{R S}{P R} \Rightarrow P R=R S \tan \theta \quad \ldots (i) \\ & \therefore P Q \cot \theta =R S \tan \theta \\ \Rightarrow & \tan \theta =\sqrt{\frac{P Q}{R S}} \quad \ldots (ii) \end{aligned}$ From Eqs. (ii) and (iii), we have $\begin{aligned} P R & =R S \sqrt{\frac{P Q}{R S}}=\sqrt{P Q \cdot R S} \\ \Rightarrow \quad 2 r & =\sqrt{P Q \cdot R S} \quad \because P R=2 r \end{aligned}$

Let \(P Q\) and \(R S\) be tangents at the extremities of a diameter \(P R\) of a circle of radius \(r\)…

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