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.

Answer 1

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.