The locus of the mid points of the chords of | vedclass

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Question

In $	riangle C P B$

$\cos \frac{	heta}{2}=\frac{P C}{2} \Rightarrow P C=2 \cos \frac{	heta}{2}$

$\Rightarrow(h-4)^2+(k-5)^2=4 \cos ^2 \frac{\

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

In $	riangle C P B$

$\cos \frac{	heta}{2}=\frac{P C}{2} \Rightarrow P C=2 \cos \frac{	heta}{2}$

$\Rightarrow(h-4)^2+(k-5)^2=4 \cos ^2 \frac{	heta}{2}$

$	ext { Now }(x-4)^2+(y-5)^2=\left(2 \cos \frac{	heta}{2}ight)^2$

$\Rightarrow I_1=2 \cos \frac{\pi}{6}=\sqrt{3}$

$I_2=2 \cos \frac{	heta_2}{2}$

$\Rightarrow I _3=2 \cos _1^2= r _2^2+ I _3^2$

$\Rightarrow 3=4 \cos ^2 \frac{	heta_2}{2}+1$

$\Rightarrow 4 \cos ^2 \frac{	heta_2}{2}=2$

$\Rightarrow \cos ^2 \frac{	heta_2}{2}=\frac{1}{2}$

$	heta_2=\frac{\pi}{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

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