- Home
- Standard 11
- Mathematics
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
$\frac{\pi}{4}$
$\frac{3 \pi}{4}$
$\frac{\pi}{6}$
$\frac{\pi}{2}$
Solution
In $\triangle C P B$
$\cos \frac{\theta}{2}=\frac{P C}{2} \Rightarrow P C=2 \cos \frac{\theta}{2}$
$\Rightarrow(h-4)^2+(k-5)^2=4 \cos ^2 \frac{\theta}{2}$
$\text { Now }(x-4)^2+(y-5)^2=\left(2 \cos \frac{\theta}{2}\right)^2$
$\Rightarrow I_1=2 \cos \frac{\pi}{6}=\sqrt{3}$
$I_2=2 \cos \frac{\theta_2}{2}$
$\Rightarrow I _3=2 \cos _1^2= r _2^2+ I _3^2$
$\Rightarrow 3=4 \cos ^2 \frac{\theta_2}{2}+1$
$\Rightarrow 4 \cos ^2 \frac{\theta_2}{2}=2$
$\Rightarrow \cos ^2 \frac{\theta_2}{2}=\frac{1}{2}$
$\theta_2=\frac{\pi}{2}$
