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)

In $\triangle CPB$

$ \cos \frac{\theta}{2}=\frac{P C}{2} \Rightarrow P C=2 \cos \frac{\theta}{2} $
$ \Rightarrow(h-4)^2+(k-5)^2=4 \cos ^2 \frac{\theta}{2} $
Now $(x-4)^2+(y-5)^2=\left(2 \cos \frac{\theta}{2}\right)^2$
$ \Rightarrow r_1=2 \cos \frac{\pi}{6}=\sqrt{3}$
$ r_2=2 \cos \frac{\theta_2}{2}$
$ r_3=2 \cos \frac{\pi}{3}=1 $
$ \Rightarrow r_1^2=r_2^2+r_3^2 $
$ \Rightarrow 3=4 \cos ^2 \frac{\theta_2}{2}+1 $
$ \Rightarrow 4 \cos ^2 \frac{\theta_2}{2}=2 $
$\Rightarrow \cos ^2 \frac{\theta_2}{2}=\frac{1}{2} $
$ \Rightarrow \theta_2=\frac{\pi}{2}$