In the figure,the pair of tangents AP and AQ | vedclass

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(B) Given that $AP$ and $AQ$ are tangents from an external point $A$ to a circle with centre $O$.Since the tangents are perpendicular to each other,$\

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$5$

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(B) Given that $AP$ and $AQ$ are tangents from an external point $A$ to a circle with centre $O$.Since the tangents are perpendicular to each other,$\angle PAQ = 90^{\circ}$.Join $OP$ and $OQ$. Also,join $OA$.In quadrilateral $APOQ$,the radius is perpendicular to the tangent at the point of contact,so $\angle OPA = 90^{\circ}$ and $\angle OQA = 90^{\circ}$.Since $\angle PAQ = 90^{\circ}$,$\angle OPA = 90^{\circ}$,and $\angle OQA = 90^{\circ}$,the fourth angle $\angle POQ$ must also be $90^{\circ}$ (as the sum of angles in a quadrilateral is $360^{\circ}$).Thus,$APOQ$ is a square because $AP = AQ$ (tangents from an external point) and all its angles are $90^{\circ}$.Therefore,the radius $OP = AP = 5 \, cm$.

In the figure,the pair of tangents $AP$ and $AQ$ drawn from an external point $A$ to a circle with centre $O$ are perpendicular to each other and the length of each tangent is $5 \, cm$. Then the radius of the circle is (in $cm$):

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