Let PQ and RS be the 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 the value of 2r is equal to

Question

Let PQ and RS be the 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 the value of 2r is equal to (A) √P Q ⋅ R S (B) (P Q + R S/2) (C) (2 P Q ⋅ R S/P Q + R S) (D) √(P Q2 + R S2/2)

Tardigrade Admin · Accepted Answer

From the figure, it is clear that Δ PRQ and Δ RSP are similar. <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-85qcujtpdgbxgxoz.jpg" /> ∴ tanθ =(P R/R S)=(P Q/R P) ⇒ PR2=PQ⋅ RS ⇒ PR=√P Q ⋅ R S ⇒ 2r=√P Q ⋅ R S

Q. Let $PQ$ and $RS$ be the 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 the value of $2 r$ is equal to

$PQ \cdot RS$

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

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

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

From the figure, it is clear that $Δ PRQ$ and $Δ RSP$ are similar.

$∴ t an θ = \frac{PR}{RS} = \frac{PQ}{RP}$
$\Rightarrow P R^{2} = PQ \cdot RS$
$\Rightarrow PR = PQ \cdot RS$
$\Rightarrow 2 r = PQ \cdot RS$

Q. Let PQ and RS be the 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 the value of 2r is equal to

PQ⋅RS​

2PQ+RS​

PQ+RS2PQ⋅RS​

2PQ2+RS2​​