The locus of the midpoints of the chords of t | vedclass

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(D) Let $C(4, 5)$ be the centre of the circle with radius $R = 2$. Let $P(h, k)$ be the midpoint of a chord $AB$ that subtends an angle $	heta$ at th

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$\frac{\pi}{2}$

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(D) Let $C(4, 5)$ be the centre of the circle with radius $R = 2$. Let $P(h, k)$ be the midpoint of a chord $AB$ that subtends an angle $	heta$ at the centre $C$.In the right-angled triangle $	riangle CPB$,we have $CP = R \cos(\frac{	heta}{2}) = 2 \cos(\frac{	heta}{2})$.The distance $CP$ is the distance from the centre $(4, 5)$ to the point $P(h, k)$,so $CP^2 = (h-4)^2 + (k-5)^2$.Thus,$(h-4)^2 + (k-5)^2 = 4 \cos^2(\frac{	heta}{2})$.This represents a circle with radius $r = 2 \cos(\frac{	heta}{2})$.Given $r_i = 2 \cos(\frac{	heta_i}{2})$,we have $r_i^2 = 4 \cos^2(\frac{	heta_i}{2})$.Given $	heta_1 = \frac{\pi}{3}$,$r_1^2 = 4 \cos^2(\frac{\pi}{6}) = 4 	imes (\frac{\sqrt{3}}{2})^2 = 3$.Given $	heta_3 = \frac{2\pi}{3}$,$r_3^2 = 4 \cos^2(\frac{\pi}{3}) = 4 	imes (\frac{1}{2})^2 = 1$.Since $r_1^2 = r_2^2 + r_3^2$,we have $3 = r_2^2 + 1$,which implies $r_2^2 = 2$.Substituting $r_2^2 = 4 \cos^2(\frac{	heta_2}{2})$,we get $4 \cos^2(\frac{	heta_2}{2}) = 2$,so $\cos^2(\frac{	heta_2}{2}) = \frac{1}{2}$.This means $\cos(\frac{	heta_2}{2}) = \frac{1}{\sqrt{2}}$,so $\frac{	heta_2}{2} = \frac{\pi}{4}$,which gives $	heta_2 = \frac{\pi}{2}$.

The locus of the midpoints 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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