A tangent from P,a point in the exterior of a | vedclass

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(B) Let $O$ be the center of the circle and $r$ be its radius.Since $PQ$ is a tangent to the circle at $Q$,the radius $OQ$ is perpendicular to the tan

Vedclass · Accepted Answer

$42$

Vedclass · Answer

(B) Let $O$ be the center of the circle and $r$ be its radius.Since $PQ$ is a tangent to the circle at $Q$,the radius $OQ$ is perpendicular to the tangent $PQ$ (i.e.,$\angle OQP = 90^\circ$).In the right-angled triangle $	riangle OQP$,by the Pythagorean theorem:$OP^2 = OQ^2 + PQ^2$Given $OP = 29$ and $PQ = 20$,we substitute these values:$29^2 = r^2 + 20^2$$841 = r^2 + 400$$r^2 = 841 - 400$$r^2 = 441$$r = \sqrt{441} = 21$.The diameter of the circle is $d = 2r = 2 	imes 21 = 42$.

$A$ tangent from $P$,a point in the exterior of a circle,touches the circle at $Q$. If $OP = 29$ and $PQ = 20$,then the diameter of the circle is:

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