Let P Q and R S be tangents at the extremities of the diameter P R of a circle of radius r . If P S and R Q intersect at a point X on the circumference of the circle, then 2 r equals [2001]

Question

Let $P Q$ and $R S$ be tangents at the extremities of the diameter $P R$ of a circle of radius $r$ . If $P S$ and $R Q$ intersect at a point $X$ on the circumference of the circle, then $2 r$ equals [2001]

Smart Achievers · Accepted Answer

(1)

$Let ∠ R P S = θ,$ $∴$ $R Q$

$and ∠ P Q X = θ$ $(∵ ∠ P X Q = 90 °)$

$∴$ $∆ P R S ~ ∆ Q P R$ $(By AA similarity)$

$∴$ $\frac{P R}{Q P} = \frac{R S}{P R} \Rightarrow P R^{2} = P Q \cdot R S$

$\Rightarrow P R = \sqrt{P Q \cdot R S} \Rightarrow 2 r = \sqrt{P Q \cdot R S}$

Solution

Q.
Let $P Q$ and $R S$ be tangents at the extremities of the diameter $P R$ of a circle of radius $r$ . If $P S$ and $R Q$ intersect at a point $X$ on the circumference of the circle, then $2 r$ equals [2001]

1 $\sqrt{P Q \cdot R S}$

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

3 $\frac{2 P Q \cdot R S}{P Q + R S}$

4 $\frac{\sqrt{(P Q^{2} + R S^{2})}}{2}$

Want Us to Email you About Special Offers & Updates?

Solution

Q. 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 2r equals [2001]

1 PQ·RS

2 PQ+RS2

3 2PQ·RSPQ+RS

4 (PQ2+RS2)2