Let C be a circle with radius 3 and center at | vedclass

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(C) Let $AB$ be a chord of the circle that subtends an angle of $\frac{2\pi}{3}$ at the center $C(0, 0)$.Let $D$ be the midpoint of the chord $AB$. Th

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

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(C) Let $AB$ be a chord of the circle that subtends an angle of $\frac{2\pi}{3}$ at the center $C(0, 0)$.Let $D$ be the midpoint of the chord $AB$. Then $CD \perp AB$.In $\Delta ACD$,the angle $\angle ACD = \frac{1}{2} \angle ACB = \frac{1}{2} 	imes \frac{2\pi}{3} = \frac{\pi}{3}$.The radius $CA = 3$.Let the coordinates of the midpoint $D$ be $(h, k)$. Then $CD = \sqrt{h^2 + k^2}$.In the right-angled triangle $\Delta ACD$,we have $\cos(\angle ACD) = \frac{CD}{CA}$.$\cos(\frac{\pi}{3}) = \frac{\sqrt{h^2 + k^2}}{3}$.$\frac{1}{2} = \frac{\sqrt{h^2 + k^2}}{3}$.$\sqrt{h^2 + k^2} = \frac{3}{2}$.Squaring both sides,we get $h^2 + k^2 = \frac{9}{4}$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = \frac{9}{4}$.

Let $C$ be a circle with radius $3$ and center at $(0, 0)$. The equation of the locus of the midpoint of the chord of the circle that subtends an angle of $\frac{2\pi}{3}$ at the center is:

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