Chapter 10 · Question 6

Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$ . Prove that $\angle PTQ = 2\angle OPQ$ .

Answer Revealed

Direct Answer:

Let

\angle PTQ = \theta

. By Theorem 10.2,

TP = TQ

, so

\triangle TPQ

is isosceles. Thus

\angle TPQ = \angle TQP = \frac{1}{2}(180^\circ - \theta) = 90^\circ - \frac{\theta}{2}

. By Theorem 10.1,

\angle OPT = 90^\circ

. Then

\angle OPQ = \angle OPT - \angle TPQ = 90^\circ - (90^\circ - \frac{\theta}{2}) = \frac{\theta}{2}

. Therefore,

\angle PTQ = \theta = 2\angle OPQ

Simple Explanation

The angle between the two tangents at the external point is always twice the angle between the radius and the chord connecting the two points of contact.

Exam-Ready Structure

Given: A circle with centre

O

, external point

T

, tangents

TP

and

TQ

touching the circle at

P

and

Q

. To Prove:

\angle PTQ = 2\angle OPQ

. Proof: Let

\angle PTQ = \theta

. By Theorem 10.2,

TP = TQ

(lengths of tangents from an external point are equal). Therefore,

\triangle TPQ

is an isosceles triangle. In

\triangle TPQ

\angle TPQ = \angle TQP = \frac{1}{2}(180^\circ - \theta) = 90^\circ - \frac{\theta}{2}

. By Theorem 10.1, the radius is perpendicular to the tangent at the point of contact, so

\angle OPT = 90^\circ

. Now,

\angle OPQ = \angle OPT - \angle TPQ = 90^\circ - \left(90^\circ - \frac{\theta}{2}\right) = \frac{\theta}{2}

. Hence,

\theta = 2\angle OPQ

, i.e.,

\angle PTQ = 2\angle OPQ

. This relationship is useful for solving angle problems involving tangents drawn from an external point.

Key Points

Let ∠PTQ = θ, then ∠TPQ = ∠TQP = 90° − θ/2
∠OPT = 90° (tangent ⟂ radius)
∠OPQ = 90° − (90° − θ/2) = θ/2
Therefore ∠PTQ = 2∠OPQ

Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$ . Prove that $\angle PTQ = 2\angle OPQ$ .