Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $\angle PTQ = 2\angle OPQ$.
Previously asked in CBSE board exam
2026 30/2/1 Q29(b); 2023 30/1/1 Q29(a) (OR-1); 2023 30/6/1 Q32(A) — asked 3×
Generated by claude-sonnet-4-6 · 2026-06-15 10:26 · grounding rag
Model Answer
Given: TP and TQ are tangents from external point T to a circle with centre O; P and Q are points of contact.
To prove: ∠PTQ = 2∠OPQ
Proof:
Let ∠PTQ = θ.
Since TP = TQ (tangents from an external point are equal), △TPQ is isosceles.
$$\angle TPQ = \angle TQP = \frac{180° - \theta}{2} = 90° - \frac{\theta}{2}$$
By Theorem 10.1, OP ⊥ TP, so ∠OPT = 90°.
$$\angle OPQ = \angle OPT - \angle TPQ = 90° - \left(90° - \frac{\theta}{2}\right) = \frac{\theta}{2}$$
$$\therefore \angle PTQ = 2\angle OPQ \qquad \textbf{(Proved)}$$
Source: Chapter 10, Section 10.3 (Example 2)
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Explanation
- Key theorems used: (1) Tangents from an external point are equal (Theorem 10.2), making △TPQ isosceles. (2) The radius is perpendicular to the tangent at the point of contact (Theorem 10.1), giving ∠OPT = 90°.
- Introduce θ for ∠PTQ to keep algebra clean — examiners appreciate this.
- The critical step is expressing ∠OPQ = ∠OPT − ∠TPQ and simplifying to θ/2.
- Write "Proved" or the QED conclusion clearly — it carries a mark.