Point P lies in the exterior of a circle with | vedclass

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(D) In a circle with centre $O$,the radius $OQ$ is perpendicular to the tangent $PQ$ at the point of contact $Q$. Therefore,$	riangle OQP$ is a right

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(D) In a circle with centre $O$,the radius $OQ$ is perpendicular to the tangent $PQ$ at the point of contact $Q$. Therefore,$	riangle OQP$ is a right-angled triangle with $\angle OQP = 90^{\circ}$.Using the Pythagorean theorem in $	riangle OQP$:$OQ^2 + PQ^2 = OP^2$Given $OP = 34$ and $PQ = 16$:$OQ^2 + 16^2 = 34^2$$OQ^2 + 256 = 1156$$OQ^2 = 1156 - 256 = 900$$OQ = \sqrt{900} = 30$.The radius of the circle is $30$. The diameter of the circle is $2 	imes 	ext{radius} = 2 	imes 30 = 60$.

Point $P$ lies in the exterior of a circle with centre $O$. $OP = 34$ and a tangent through $P$ touches the circle at $Q$. If $PQ = 16$,then find the diameter of the circle.

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