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Q.

Let PQ and RS be tangents at the extremities of a diameter PR of a circle of radius r such that PS and RQ intersect at a point X on the circumference of the circle, then 2r equals

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a

$\frac{P Q + R S}{2}$

b

$\sqrt{\frac{{(P Q)}^{2} + {(R S)}^{2}}{2}}$

c

$\sqrt{P Q . R S}$

d

$\frac{2 P Q . R S}{P Q + R S}$

answer is A.

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Detailed Solution

$\begin{aligned} \tan θ = \frac{P Q}{P R} = \frac{P Q}{2 r} \\ also \tan (\frac{π}{2} - θ) = \frac{R S}{2 r} \end{aligned}$
$\begin{aligned} \cot θ = \frac{R S}{2 r} \\ ∴ \tan θ \cot θ = \frac{P Q \cdot R S}{4 r^{2}} \\ \Rightarrow 4 r^{2} = P Q \cdot R S \\ \Rightarrow 2 r = \sqrt{P Q \cdot R S} \end{aligned}$

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