Let C be the circle with centre (0,0) and rad | vedclass

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(C) Let the coordinates of a point $P$ be $(h, k)$,which is the midpoint of the chord $AB$.Now,$OP = \sqrt{(h-0)^2 + (k-0)^2} = \sqrt{h^2+k^2}$.In $

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$x^2+y^2=\frac{9}{4}$

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(C) Let the coordinates of a point $P$ be $(h, k)$,which is the midpoint of the chord $AB$.Now,$OP = \sqrt{(h-0)^2 + (k-0)^2} = \sqrt{h^2+k^2}$.In $	riangle AOP$,the angle $\angle AOP = \frac{1}{2} 	imes \frac{2\pi}{3} = \frac{\pi}{3}$.Using trigonometry in $	riangle AOP$,we have $\cos\left(\frac{\pi}{3}ight) = \frac{OP}{OA}$.Since $OA$ is the radius of the circle,$OA = 3$.$\Rightarrow \frac{1}{2} = \frac{\sqrt{h^2+k^2}}{3}$.$\Rightarrow \sqrt{h^2+k^2} = \frac{3}{2}$.Squaring both sides,we get $h^2+k^2 = \frac{9}{4}$.Replacing $(h, k)$ with $(x, y)$,the required locus is $x^2+y^2 = \frac{9}{4}$.

Let $C$ be the circle with centre $(0,0)$ and radius $3$ units. The equation of the locus of the midpoints of the chords of the circle $C$ that subtend an angle of $\frac{2\pi}{3}$ at its centre is:

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