Mathematics — CBSE Class 10 board question

Given: Circle with centre O, radius OP = OQ = 5 cm, chord PQ = 8 cm. Tangents at P and Q meet at T.

Step 1: Join OT. Since TP = TQ (tangents from external point are equal), △TPQ is isosceles and OT is the angle bisector of ∠PTQ. Therefore, OT ⊥ PQ and bisects PQ.

So, PR = RQ = 4 cm (where R is the point where OT meets PQ).

Step 2: In △ORP (right-angled at R):
$$OR = \sqrt{OP^2 - PR^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \text{ cm}$$

Step 3: Let TP = $x$ and TR = $y$.

From △TRP: $x^2 = y^2 + 16$ … (1)

From △OPT: $x^2 + 25 = (y + 3)^2$ … (2)

Subtracting (1) from (2):
$$25 = 6y - 7 \implies y = \frac{16}{3}$$

Substituting in (1):
$$x^2 = \frac{256}{9} + 16 = \frac{256 + 144}{9} = \frac{400}{9}$$

$$\therefore TP = x = \frac{20}{3} \text{ cm}$$

Source: Chapter 10, Section 10.3 (Example 3)

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• The key properties used are: (1) tangent ⊥ radius at point of contact, and (2) tangents from an external point are equal in length.
• Since TP = TQ, OT acts as the perpendicular bisector of chord PQ — this is the crucial geometric insight.
• Examiners award marks for each step: finding PR, finding OR, setting up the two Pythagoras equations, and the final answer $\frac{20}{3}$ cm.
• Alternatively, similarity of △TRP and △PRO gives the answer more elegantly: $\frac{TP}{OP} = \frac{RP}{OR} \Rightarrow TP = \frac{4 \times 5}{3} = \frac{20}{3}$ cm. Either method is accepted.