P is in the exterior of odot(O, 30). The tang | vedclass

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(D) Given that $O$ is the center of the circle and $r = 30$ is the radius.Since $PQ$ is a tangent to the circle at point $Q$,the radius $OQ$ is perpen

Vedclass · Accepted Answer

$16$

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(D) Given that $O$ is the center of the circle and $r = 30$ is the radius.Since $PQ$ is a tangent to the circle at point $Q$,the radius $OQ$ is perpendicular to the tangent $PQ$ at the point of contact.Therefore,$	riangle OQP$ is a right-angled triangle with $\angle OQP = 90^{\circ}$.In $	riangle OQP$,by the Pythagorean theorem:$OP^2 = OQ^2 + PQ^2$Given $OP = 34$ and $OQ = r = 30$.$34^2 = 30^2 + PQ^2$$1156 = 900 + PQ^2$$PQ^2 = 1156 - 900$$PQ^2 = 256$$PQ = \sqrt{256} = 16$.Thus,the length of the tangent $PQ$ is $16$.

$P$ is in the exterior of $\odot(O, 30)$. The tangent drawn from $P$ to the circle touches the circle at $Q$. If $OP = 34$,then $PQ = \dots$

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