odot(P, 3) and odot(P, 5) are two concentric | vedclass

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(C) Chord $\overline{AB}$ of $\odot(P, 5)$ touches $\odot(P, 3)$ at $M$.Since the radius is perpendicular to the tangent at the point of contact,$\ove

Vedclass · Accepted Answer

$8$

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(C) Chord $\overline{AB}$ of $\odot(P, 5)$ touches $\odot(P, 3)$ at $M$.Since the radius is perpendicular to the tangent at the point of contact,$\overline{PM} \perp \overline{AB}$.In a circle,a perpendicular drawn from the center to a chord bisects the chord. Therefore,$M$ is the midpoint of $\overline{AB}$.Given $PM = 3$ (radius of the smaller circle) and $PB = 5$ (radius of the larger circle).In right-angled $\Delta PMB$,by Pythagoras theorem:$PB^2 = PM^2 + MB^2$$5^2 = 3^2 + MB^2$$25 = 9 + MB^2$$MB^2 = 25 - 9 = 16$$MB = \sqrt{16} = 4$.Since $M$ is the midpoint of $\overline{AB}$,$AB = 2 	imes MB = 2 	imes 4 = 8$.

$\odot(P, 3)$ and $\odot(P, 5)$ are two concentric circles. Chord $\overline{AB}$ of $\odot(P, 5)$ touches $\odot(P, 3)$ at $M$. Find $AB$.

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