Mathematics — Standard (041) — CBSE Class 10 board question

Given: A circle with centre O, an external point P, and two tangents PQ and PR to the circle at points Q and R respectively.

To prove: PQ = PR

Construction: Join OP, OQ, and OR.

Proof:

Since OQ and OR are radii and PQ, PR are tangents at Q and R,

$$\angle OQP = \angle ORP = 90° \quad \text{(Radius ⊥ tangent at point of contact)}$$

In right triangles OQP and ORP:

| Statement | Reason |
|---|---|
| OQ = OR | Radii of same circle |
| OP = OP | Common |
| ∠OQP = ∠ORP = 90° | Radius ⊥ tangent |

∴ △OQP ≅ △ORP (RHS congruence)

∴ PQ = PR (CPCT) $\hspace{2cm}$ Proved

Source: Theorem 10.2, Chapter 10 — Circles

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• Examiners expect: Given, To prove, Construction, Proof — all four parts for full marks.
• The key steps are: angles at Q and R are 90° (from Theorem 10.1), then RHS congruence, then CPCT.
• You may also use the Pythagoras alternative: $PQ^2 = OP^2 - OQ^2 = OP^2 - OR^2 = PR^2$, so PQ = PR. Either method earns full marks.
• Don't skip stating "radius ⊥ tangent" as the reason for 90°; examiners specifically look for it.