PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (see Fig. 10.10). Find the length TP.
Step-by-Step Solution
Key Concept: Use the relationship between the chord length and the central angle ( $PQ = 2R\sin\frac{\theta}{2}$ ), then apply the law of cosines in the isosceles triangle $\triangle PTQ$ formed by the two tangents. The tangents are equal, so $TP = TQ$, and the angle $\angle PTQ = 180^{\circ}-\theta$. This gives a solvable equation for $TP$.
1. Let $O$ be the centre of the circle and $\theta = \angle POQ$ be the central angle subtended by chord $PQ$.
Using the chord‑radius formula:
$$PQ = 2R\sin\frac{\theta}{2}$$
Substituting $PQ = 8\,\text{cm}$ and $R = 5\,\text{cm}$,
$$8 = 2\times5\sin\frac{\theta}{2}\;\Rightarrow\;\sin\frac{\theta}{2}=\frac{8}{10}=0.8.$$
Hence
$$\cos\frac{\theta}{2}=\sqrt{1-\sin^{2}\frac{\theta}{2}}=\sqrt{1-0.64}=0.6.$$
Using $\cos\theta = 2\cos^{2}\frac{\theta}{2}-1$,
$$\cos\theta = 2(0.6)^{2}-1 = 0.72-1 = -0.28.$$
2. The tangents at $P$ and $Q$ meet at $T$. Since tangents from an external point are equal, let
$$TP = TQ = x\;\text{cm}.$$
The angle between the two tangents is
$$\angle PTQ = 180^{\circ}-\theta.$$
In $\triangle PTQ$ (isosceles with sides $x, x, 8$) apply the law of cosines:
$$\begin{aligned}
PQ^{2} &= x^{2}+x^{2}-2x^{2}\cos(\angle PTQ)\\
8^{2} &= 2x^{2}\bigl[1-\cos(180^{\circ}-\theta)\bigr]\\
64 &= 2x^{2}\bigl[1+\cos\theta\bigr] \quad\text{(since }\cos(180^{\circ}-\theta)=-\cos\theta\text{)}.
\end{aligned}$$
Substituting $\cos\theta = -0.28$ gives
$$64 = 2x^{2}(1-0.28)=2x^{2}(0.72)\;\Rightarrow\;x^{2}=\frac{64}{1.44}=\frac{400}{9}.$$
Hence
$$x = \sqrt{\frac{400}{9}} = \frac{20}{3}\;\text{cm}.$$
3. Therefore
$$TP = \frac{20}{3}\,\text{cm} \approx 6.67\,\text{cm}.$$
(An alternative check: using the power of a point, $TP^{2}=OT^{2}-R^{2}$, and $OT = OM+MT = 3+\frac{16}{3}=\frac{25}{3}$, which also yields $TP = \frac{20}{3}$ cm.)
Correct Answer: TP = \frac{20}{3}\ \text{cm}