JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Circle

Consider two circles

C_{1} : x^{2} + y^{2} = 25

and

C_{2} : (x - α)^{2} + y^{2} = 16

, where

α \in (5, 9)

. Let the angle between the two radii (one to each circle) drawn from one of the intersection points of

C_{1}

and

C_{2}

sin^{- 1} (\frac{63}{8})

. If the length of common chord of

C_{1}

and

C_{2}

β

, then the value of

(α β)^{2}

equals

Answer: 1575

Solution

C_{1}

has center

O

, radius 5;

C_{2}

has center

P = (α, 0)

, radius 4. At an intersection point

A

O A = 5

P A = 4

, and the angle

∠ O A P = θ

with

sin θ = \frac{63}{8}

.

The area of triangle

O A P

can be written two ways.

Using the included angle at

A

: Area

= \frac{1}{2} \cdot O A \cdot P A \cdot sin θ = \frac{1}{2} \cdot 5 \cdot 4 \cdot \frac{63}{8} = \frac{5 63}{4}

.

Using base

O P = α

and height equal to half the common chord (the two intersection points are symmetric about line

O P

, each at perpendicular distance

\frac{β}{2}

from it): Area

= \frac{1}{2} \cdot α \cdot \frac{β}{2} = \frac{α β}{4}

.

Equating:

\frac{α β}{4} = \frac{5 63}{4} \Rightarrow α β = 563

(α β)^{2} = 25 \times 63 = 1575

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Integer answer