overline{PA} is a tangent to odot(O, r) drawn | vedclass

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(C) Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$\angle OAP = 90^{\circ}$.In right-angled t

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$12$

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(C) Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$\angle OAP = 90^{\circ}$.In right-angled triangle $	riangle OAP$,by Pythagoras theorem:$OP^{2} = OA^{2} + AP^{2}$Given $OP = 10$ and $AP = 8$,we have:$10^{2} = OA^{2} + 8^{2}$$100 = OA^{2} + 64$$OA^{2} = 100 - 64 = 36$$OA = \sqrt{36} = 6$Here,$OA$ is the radius $(r)$ of the circle.Diameter of the circle $= 2 	imes 	ext{radius} = 2 	imes 6 = 12$.

$\overline{PA}$ is a tangent to $\odot(O, r)$ drawn from a point $P$ outside a circle. If $OP = 10$ and $AP = 8$,then the diameter of the circle is equal to $\ldots \ldots \ldots \ldots .$

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