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

Given: Two circles with centres O and O', radii 6 cm and 8 cm, intersecting at P and Q. OP is tangent to the circle with centre O', and O'P is tangent to the circle with centre O.

Step 1: Find OO'

Since OP ⊥ O'P (tangent ⊥ radius at point of contact, by Theorem 10.1):
$$OO'^2 = OP^2 + O'P^2 = 6^2 + 8^2 = 36 + 64 = 100$$
$$OO' = 10 \text{ cm}$$

Step 2: Find area of △OPO'

$$\text{Area} = \frac{1}{2} \times OP \times O'P = \frac{1}{2} \times 6 \times 8 = 24 \text{ cm}^2$$

Step 3: Let PL ⊥ OO', where L lies on OO'

Also, Area $= \frac{1}{2} \times OO' \times PL$

$$24 = \frac{1}{2} \times 10 \times PL \implies PL = \frac{48}{10} = 4.8 \text{ cm}$$

Step 4: PQ = 2 × PL (since OO' is the perpendicular bisector of common chord PQ)

$$PQ = 2 \times 4.8 = \boxed{9.6 \text{ cm}}$$

Source: Chapter 10, Section 10.2 (Theorem 10.1)

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• The key insight is that since OP is tangent to circle O' and O'P is tangent to circle O, by Theorem 10.1 each radius is perpendicular to the tangent at the point of contact — making ∠OPO' = 90°.
• Examiners expect you to use the area method: equate two expressions for the area of △OPO' to find the perpendicular PL from P to OO'.
• Since the line joining two centres OO' is the perpendicular bisector of the common chord PQ, you double PL to get PQ.
• Don't forget to state the theorem (tangent ⊥ radius) — it earns a dedicated mark.