Question

Two circles touch each other externally at P. AB is a common tangent to the circle touching them at A and B. The value of $\angle$APB is

(a) 30º (b) 45º (c) 60º (d) 90º [CBSE 2014]

Answer 1

It is given that two circles touch each other externally at P. AB is a common tangent to the circles touching them at A and B.



Draw a tangent to the circles at P, intersecting AB at T.

Now, TA and TP are tangents drawn to the same circle from an external point T.

∴ TA = TP (Lengths of tangents drawn from an external point to a circle are equal)

TB and TP are tangents drawn to the same circle from an external point T.

∴ TB = TP (Lengths of tangents drawn from an external point to a circle are equal)

In ∆ATP,

TA = TP

$\therefore \angle \mathrm{APT}=\angle \mathrm{PAT}$ .....(1) (In a triangle, equal sides have equal angles opposite to them)

In ∆BTP,

TB = TP

$\therefore \angle \mathrm{BPT}=\angle \mathrm{PBT}$ .....(2) (In a triangle, equal sides have equal angles opposite to them)

Now, in ∆APB,

$\angle \mathrm{APB}+\angle \mathrm{PAB}+\angle \mathrm{PBA}=180°$ (Angle sum property)
$$\begin{gathered} \Rightarrow \angle \mathrm{APB}+\angle \mathrm{APT}+\angle \mathrm{BPT}=180°\left[\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\right] \\ \Rightarrow \angle \mathrm{APB}+\angle \mathrm{APB}=180° \\ \Rightarrow 2\angle \mathrm{APB}=180° \\ \Rightarrow \angle \mathrm{APB}=90° \end{gathered}$$

Thus, the value of $\angle$APB is 90º.

Hence, the correct answer is option D.