Two circles touch each other externally at P. AB is a common tang…

MCQ

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 $\angle\text{APB}$ is :

A
$30^\circ$
B
$45^\circ$
C
$60^\circ$
✓
$90^\circ$

✓

Answer

Correct option: D.

$90^\circ$

It is given that two circles touch each other externally at $P.$
$ AB$ is a common tangent to the circle touching them at $A$ and $B$.
Draw a tangent to the circle at $P,$ intersecting $AB$ at $T.$
Now, $TA$ and $TP$ are tangent drawn to the same circle from an external point $T.$
$\therefore TA = TP \ ($Length of tangents drawn from an external point to a circle are equal$)$
$TB$ and $TP$ are tangent drawn to the same circle from an external point $T$.
$\therefore TB = TP \ ($Length of tangents drawn from an external point to a circle are equal$)$
In $\triangle\text{ATP}$
$TA = TP$
$\therefore\angle\text{APT}=\angle\text{PAT}...(1)\ ($In a triangle, equal sides have equal angles opposite to them$)$
In $\triangle\text{BTP},$
$TB = TP$
$\therefore\angle\text{BPT}=\angle\text{PBT}...(2)\ ($In a triangle, equal sides have equal angles opposite to them$)$
Now, in $\triangle\text{APB},$
$\Rightarrow\angle\text{APB}+\angle\text{PAB}+\angle\text{PBA}=180^\circ\ ($Angle sum property$)$
$\Rightarrow\angle\text{APB}+\angle\text{APT}+\angle\text{BPT}=180^\circ\ [$From $(1)$ and $(2)]$
$\Rightarrow\angle\text{APB}+\angle\text{APB}=180^\circ$
$\Rightarrow2\angle\text{APB}=180^\circ$
$\Rightarrow\angle\text{APB}=90^\circ$
Thus, the value of $\angle\text{APB}\ \text{is}\ 90^\circ$