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Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact to the centre.

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Question

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact to the centre.

Theorem
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Solution

Let us consider a circle centered at point O. Let P be an external point from which two tangents PA and PB are drawn to the circle, which are touching the circle at points A and B respectively and AB is the line segment joining points of contact A and B together such that it subtends ∠AOB at center O of the circle.

It can be observed that

OA (radius) ⊥ PA (tangent)

Therefore, ∠OAP = 90°

Similarly, OB (radius) ⊥ PB (tangent)

∠OBP = 90°

In quadrilateral OAPB,

Sum of all interior angles = 360º

∠OAP + ∠APB + ∠PBO + ∠BOA = 360º

90º + ∠APB + 90º + ∠BOA = 360º

∠APB + ∠BOA = 180º

Hence, it can be observed that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

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Chapter 10: Circles - EXERCISE 10.2 [Page 152]
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