Question

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

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{6}$
• D. $\frac{3 \pi}{4}$

Answer 1

Answer: (A)

$\frac{\pi}{2}$

Answer 2

The correct answer is (A) : $\frac{\pi}{2}$

In △CPB

cos2θ​=2PC​⇒PC=2cos2θ​
⇒(h−4)2+(k−5)2=4cos22θ​
Now (x−4)2+(y−5)2=(2cos2θ​)2
⇒r1​=2cos6π​=3​
r2​=2cos2θ2​​
r3​=2cos3π​=1
⇒r12​=r22​+r32​
⇒3=4cos22θ2​​+1
⇒4cos22θ2​​=2
⇒cos22θ2​​=21​
⇒θ2​=2π​