The chord of odot(O, 34) touches odot(O, 16). | vedclass

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(B) Let the two concentric circles be $C_1$ with radius $R = 34$ and $C_2$ with radius $r = 16$.Let $AB$ be the chord of the larger circle $C_1$ that

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$60$

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(B) Let the two concentric circles be $C_1$ with radius $R = 34$ and $C_2$ with radius $r = 16$.Let $AB$ be the chord of the larger circle $C_1$ that touches the smaller circle $C_2$ at point $P$.Since $AB$ is tangent to $C_2$ at $P$,the radius $OP$ is perpendicular to $AB$ $(OP \perp AB)$.In the right-angled triangle $	riangle OPA$,by the Pythagorean theorem:$OA^2 = OP^2 + AP^2$$34^2 = 16^2 + AP^2$$1156 = 256 + AP^2$$AP^2 = 1156 - 256 = 900$$AP = \sqrt{900} = 30$.Since the perpendicular from the center to a chord bisects the chord,$AB = 2 	imes AP$.$AB = 2 	imes 30 = 60$.

The chord of $\odot(O, 34)$ touches $\odot(O, 16)$. The length of the chord is .........

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