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Question

PA and PB are tangents from P to the circle with center O. At point M which lies on circle, a tangent is drawn cutting PA at K and PB at N. Prove that KN = AK + BN. [2 MARKS]


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Solution

Concept: 1 Mark
Application: 1 Mark

We know that the tangents drawn from an external point to a circle are equal in length.

KA = KM .... (i) [From K]

and, NB = NM .... (ii) [From N]

Adding equations (i) and (ii), we get

KA + NB = KM + NM

AK+BN=KM+MN

AK+BN=KN


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