The locus of the mid points of the chords of the circle $C_{1}:(x-4)^{2}+(y-5)^{2}=4$ which subtend an angle $\theta_{i}$ at the centre of the circle $C_{1}$, is a circle of radius $r_{i}$. If $\theta_{1}=\frac{\pi}{3}, \theta_{3}=\frac{2 \pi}{3}$ and $r_{1}^{2}=r_{2}^{2}+r_{3}^{2}$, then $\theta_{2}$ is equal to

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Circle

The locus of the mid points of the chords of the circle

C_{1} : (x - 4)^{2} + (y - 5)^{2} = 4

which subtend an angle

θ_{i}

at the centre of the circle

C_{1}

, is a circle of radius

r_{i}

. If

θ_{1} = \frac{π}{3}, θ_{3} = \frac{2 π}{3}

and

r_{1}^{2} = r_{2}^{2} + r_{3}^{2}

, then

θ_{2}

is equal to

Answer: (D)

The circle

C_{1}

has centre

(4, 5)

and radius

2

. A chord subtending an angle

θ

at the centre has its midpoint at distance

2 cos \frac{θ}{2}

from the centre, since the centre, the midpoint and an endpoint form a right triangle with the radius

2

as hypotenuse. As the chord varies, its midpoint traces a circle concentric with

C_{1}

whose radius is

r = 2 cos \frac{θ}{2}

. Thus

r_{1} = 2 cos \frac{π}{6} = 3

and

r_{3} = 2 cos \frac{π}{3} = 1

, so

r_{1}^{2} = 3

and

r_{3}^{2} = 1

. The relation

r_{1}^{2} = r_{2}^{2} + r_{3}^{2}

becomes

3 = 4 cos^{2} \frac{θ _{2}}{2} + 1

, giving

cos^{2} \frac{θ _{2}}{2} = \frac{1}{2}

and

cos \frac{θ _{2}}{2} = \frac{1}{2}

, so

\frac{θ _{2}}{2} = \frac{π}{4}

. The required angle is

\frac{π}{2}

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Single correct option