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 at the centre.

Answer 1

Let us Consider a circle with centre 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 point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.
It can be observed that
OA ⊥ PA
∴ ∠OAP = 90°
Similarly, OB ⊥ PB
∴ ∠OBP = 90°
In quadrilateral OAPB,
Sum of all interior angles = 360º
∠OAP +∠APB +∠PBO +∠BOA = 360º
⇒ 90º + ∠APB + 90º + ∠BOA = 360º
⇒ ∠APB + ∠BOA = 180º

∴ 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.

Answer 2

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 point A and B respectively and AB is the line segment, joining point of contacts 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.