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

Given: A circle with centre ‘O’. 

Two tangents \(\overleftrightarrow{PQ}\)and \(\overleftrightarrow{PT}\) from an external point P. 

Let Q, T be the points of contact. 

R.T.P: ∠P and ∠QOT are supplementary. 

Proof: OQ ⊥ PQ 

[∵ radius is perpendicular to the tangent at the point of contact] also OT ⊥ PT 

∴ ∠OQP + ∠OTP = 90° + 90° = 180° 

Now in □PQOT,

∠OTP + ∠TPQ + ∠PQO + ∠QOT = 360° (angle sum property) 

180° + ∠P + ∠QOT = 360° 

∠P + ∠QOT = 360°- 180° = 180° 

Hence proved. (Q.E.D.)