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 segments joining the points of contact to the centre.

Answer 1

Given: PA and PB are the tangent drawn from a point P to a circle with centre O. Also, the line segments OA and OB are drawn.

To Prove: ∠APB + ∠AOB = 180°

Proof: We know that the tangent to a circle is perpendicular to the radius through the point of contact.

Now, in right ΔOAP and right ΔOBP, we have PA = PB     ...[Tangents to circle from an external point]

OA = OB    ...[Radii of the same circle]

OP = OP       ....[Common]

 ΔOAP ≅ ΔOBP             ....[by SSS congrurncy]

∠OPA = ∠OPB

 and ∠AOP = ∠BOP

⇒ ∠APB = 2∠OPA and ∠AOB = 2∠AOP

In right  ΔOAP

⇒ ∠AOP + ∠OPA + ∠PAO = 180°

⇒ ∠AOP = 180° − 90° − ∠OPA

⇒ ∠AOP = 90° − ∠OPA

⇒ 2∠AOP = 180° − 2∠OPA

⇒ ∠AOB = 180° − ∠APB

⇒ ∠AOB + ∠APB = 180°

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.