Answer :

Given P lies in the exterior of a circle having centre O and PQ is a tangent.

∴ OQ ⊥ PQ

Also given OP = 25 and OQ = 24.

Consider ΔOQP,

∠OQP = 90°

By Pythagoras Theorem,

⇒ OP^{2} = OQ^{2} + PQ^{2}

⇒ 25^{2} = 24^{2} + PQ^{2}

⇒ PQ^{2} = 625 – 576

= 49

∴ PQ = 7

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