In figure 3.83, M is the centre of the circle and seg KL is a tangent segment.
If MK = 12, KL =
then find -
(1) Radius of the circle.
(2) Measures of ∠ K and ∠ M.

(1) Here LM is the radius of the circle
⇒ ∠KML = 90° Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.
In triangle MLK right-angled at L,
Given MK = 12, KL = 6√3,
MK2 = LM2 + KL2 {Using Pythagoras theorem}
⇒LM2 = 122 -6√32
⇒LM2 = 144 - 108
⇒ LM = √36
⇒ LM = 6 cm
(2) ![]()
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⇒∠K = 30°
∠ M + ∠ K + ∠ L = 180° {Angle sum property of the triangle}
So, ∠ M = 180° - ∠ K - ∠ L
⇒ ∠ M = 180° - 30° - 90° = 60°
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