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 $△CPB$

$\cos \frac{\theta }{2}=\frac{PC}{2}\Rightarrow PC=2\cos \frac{\theta }{2}$
$\Rightarrow (h-4{)}^2+(k-5{)}^2=4{\cos }^2\frac{\theta }{2}$
Now $(x-4{)}^2+(y-5{)}^2={\left(2\cos \frac{\theta }{2}\right)}^2$
$\Rightarrow r_1=2\cos \frac{\pi }{6}=\sqrt{3}$
$r_2=2\cos \frac{{\theta }_2}{2}$
$r_3=2\cos \frac{\pi }{3}=1$
$\Rightarrow r_1^2=r_2^2+r_3^2$
$\Rightarrow 3=4{\cos }^2\frac{{\theta }_2}{2}+1$
$\Rightarrow 4{\cos }^2\frac{{\theta }_2}{2}=2$
$\Rightarrow {\cos }^2\frac{{\theta }_2}{2}=\frac{1}{2}$
$\Rightarrow {\theta }_2=\frac{\pi }{2}$