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 is:

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

MCQ

Two circles touch each other externally at $\text{P. AB}$ 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 $\text{P. AB}$ is a common tangent to the circle touching them at $A$ and $B$.
Draw a tangent to the circle at $P,$ intersecting $\text{AB}$ at $T.$
Now, $\text{TA}$ and $\text{TP}$ are tangent drawn to the same circle from an external point $T.$
$\therefore \text{TA = TP} ($Length of tangents drawn from an external point to a circle are equal$)$
$\text{TB}$ and $\text{TP}$ are tangent drawn to the same circle from an external point $T.$
$\therefore \text{TB = TP} ($Length of tangents drawn from an external point to a circle are equal$)$
In $\triangle\text{ATP}$
$\text{TA = TP}$
$\therefore\angle\text{APT}=\angle\text{PAT}...(1)($In a triangle, equal sides have equal angles opposite to them$)$
In $\triangle\text{BTP},$
$\text{TB = TP}$
$\therefore\angle\text{BPT}=\angle\text{PBT}...(2)($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 $\angle\text{APB}$ is $90^\circ$