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Question
In the given figure, M is the centre of the circle and seg KL is a tangent segment. L is a point of contact. If MK = 12, KL = `6sqrt3`, then find the radius of the circle.
Solution
Line KL is the tangent to the circle at point L and seg ML is the radius.
∴ ∠MLK = 90° ......…[Tangent theorem]
In ΔMLK, ∠MLK = 90°
∴ MK2 = ML2 + KL2 .....…[Pythagoras theorem]
∴ 122 = ML2 + `(6sqrt3)^2`
∴ 144 = ML2 + `36 xx 3`
∴ 144 = ML2 + 108
∴ ML2 = 144 − 108
∴ ML2 = 36
∴ ML = `sqrt36` = 6 units …[Taking square root of both sides]
∴ Radius of the circle is 6 units.
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