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

Question

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 (A) √P Q ⋅ R S (B) (P Q+R S/2) (C) (2 P Q ⋅ R S/P Q+R S) (D) (√(P Q)2+(R S)2/2)

Tardigrade Admin · Accepted Answer

According to the question, from the diagram, in

△ PQR

tan θ = \frac{PQ}{PR}

\Rightarrow PR = PQ cot θ \dots

(i)

and in

△ PRS

,

tan (9 0^{\circ} - θ) = \frac{RS}{PR}

\Rightarrow PR = RS tan θ

∴ PQ cot θ = RS tan θ

\Rightarrow tan θ = \frac{PQ}{RS} ... (iii)

From Eqs. (ii) and (iii), we have

PR = RS \frac{PQ}{RS} = PQ \cdot RS

\Rightarrow 2 r = PQ \cdot RS ∵ PR = 2 r

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 $2 r$ equals

$PQ \cdot RS$

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

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

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

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

PQ⋅RS​

2PQ+RS​

PQ+RS2PQ⋅RS​

2(PQ)2+(RS)2​​

According to the question, from the diagram, in △PQR tanθ=PRPQ​ ⇒PR=PQcotθ… (i) and in △PRS, tan(90∘−θ)=PRRS​ ⇒PR=RStanθ ∴PQcotθ=RStanθ ⇒tanθ=RSPQ​​...(iii) From Eqs. (ii) and (iii), we have PR=RSRSPQ​​=PQ⋅RS​ ⇒2r=PQ⋅RS​∵PR=2r

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 $2 r$ equals

$PQ \cdot RS$

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

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

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