Let AB be a chord of a circle and C divides A | vedclass

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(B) Let $AC = 3x$ and $CB = x$. Then $AB = 4x$.By the Power of a Point theorem at $C$,$AC \cdot CB = CD \cdot CE$.Let $h_1 = 3$ and $h_2 = 2$ be the p

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$2\sqrt{2}\pi$

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(B) Let $AC = 3x$ and $CB = x$. Then $AB = 4x$.By the Power of a Point theorem at $C$,$AC \cdot CB = CD \cdot CE$.Let $h_1 = 3$ and $h_2 = 2$ be the perpendicular distances from $D$ and $E$ to $AB$.In $	riangle D C B'$ (where $B'$ is the projection of $D$ on $AB$),$CD = \frac{h_1}{\sin \alpha} = \frac{3}{\sin \alpha}$.In $	riangle E C B''$ (where $B''$ is the projection of $E$ on $AB$),$CE = \frac{h_2}{\sin \alpha} = \frac{2}{\sin \alpha}$.Thus,$(3x)(x) = \left(\frac{3}{\sin \alpha}ight) \left(\frac{2}{\sin \alpha}ight) \implies 3x^2 = \frac{6}{\sin^2 \alpha} \implies x^2 = \frac{2}{\sin^2 \alpha} \implies x = \frac{\sqrt{2}}{\sin \alpha}$.Then $AB = 4x = \frac{4\sqrt{2}}{\sin \alpha}$.For $AB$ to be minimum,$\sin \alpha$ must be maximum,so $\sin \alpha = 1$,which means $\alpha = \frac{\pi}{2}$.Then $r = AB_{\min} = 4\sqrt{2}$.The value of $r\alpha = (4\sqrt{2}) \cdot \frac{\pi}{2} = 2\sqrt{2}\pi$.

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