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

$\Rightarrow A K + B N = K M + M N$

$\Rightarrow A K + B N = K N$

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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]

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