Circle Question 1
Question 1 - 24 January - Shift 2
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
(1) $\frac{\pi}{4}$
(2) $\frac{3 \pi}{4}$
(3) $\frac{\pi}{6}$
(4) $\frac{\pi}{2}$
Show Answer
Answer: (4)
Solution:
Formula: Distance Formula
In $\triangle CPB$
$\cos \frac{\theta}{2}=\frac{P C}{2} \Rightarrow PC=2 \cos \frac{\theta}{2}$
$\Rightarrow(h-4)^{2}+(k-5)^{2}=4 \cos ^{2} \frac{\theta}{2}$
Now $(x-4)^{2}+(y-5)^{2}=(2 \cos \frac{\theta}{2})^{2}$
$\Rightarrow r_1=2 \cos \frac{\pi}{6}=\sqrt{3}$
$\begin{aligned} & \mathrm{r}_2=2 \cos \frac{\theta_2}{2} \\ & \mathrm{r}_3=2 \cos \frac{\pi}{3}=1 \\ & \Rightarrow \mathrm{r}_1^2=\mathrm{r}_2^2+\mathrm{r}_3^2 \\ & \Rightarrow 3=4 \cos ^2 \frac{\theta_2}{2}+1 \\ & \Rightarrow 4 \cos ^2 \frac{\theta_2}{2}=2 \\ & \Rightarrow \cos ^2 \frac{\theta_2}{2}=\frac{1}{2} \\ & \Rightarrow \theta_2=\frac{\pi}{2}\end{aligned}$
