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

Question

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 (A) √PQ ⋅ RS (B)  (PQ+RS/2) (C)  (2PQ ⋅ RS/PQ+RS) (D) √ (PQ2+RS2/2)

Tardigrade Admin · Accepted Answer

From figure, it is clear that triangle PRQ and triangle RSP are similar, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/313961913dc5dd622d78513e629ce8a8-.png" /> ∴ (PR/RS)= (PQ/RP)⇒ PR2=PQ . RS ⇒ PR=√PQ ⋅ RS ⇒ 2r=√PQ ⋅ RS

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 $2 r$ equals

$PQ \cdot RS$

$\frac{PQ + RS}{2}$

$\frac{2 PQ \cdot RS}{PQ + RS}$

$\frac{PQ ^{2} + RS ^{2}}{2}$

From figure, it is clear that $△ PRQ$ and $△ RSP$ are similar,

$∴ \frac{PR}{RS} = \frac{PQ}{RP} \Rightarrow PR^{2} = PQ . RS$
$\Rightarrow PR = PQ \cdot RS$
$\Rightarrow 2 r = PQ \cdot RS$

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

PQ⋅RS​

2PQ+RS​

PQ+RS2PQ⋅RS​

2PQ2+RS2​​