Question

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

• A. $30°$
• B. $45°$
• C. $60°$
• D. $90°$

Answer 1

The correct option is D

$90°$

Finding the value of angle formed at $APB$:

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 $T$.

$\therefore TA=TP$ (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 $△ATP$,

$TA=TP$

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

In $△BTP,$

$TB=TP$

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

Now, in $△APB,$

$\angle APB+\angle PAB+\angle PBA=180°$ (Angle sum property)

$\angle APB+\angle APT+\angle BPT=180°$ [From ($1$) and ($2$)]

$\Rightarrow \angle APB+\angle APB=180°$

$$\begin{gathered} \Rightarrow 2\angle APB=180° \\ \Rightarrow \angle APB=90° \end{gathered}$$

Therefore, the value of $\angle APB$ is $90°$.