Question

Two circles touch each other externally at a point A and PQ is a common tangent which touches the circle at P and Q respectively
then $\angle PAQ$ is--

• A. ${45}^{\circ }$
• B. ${90}^{\circ }$
• C. ${80}^{\circ }$
• D. ${100}^{\circ }$

Answer 1

The correct option is B ${90}^{\circ }$

$\because$ Two circles with centres $O_1$ and $O_2$
touch each other externally at a point A and
PQ is a common tangent which touches the
circle at the points P and Q respectively
Let $\angle PAQ=a^{\circ }$
and $\angle O_{1}AP=\angle O_{1}PA={\theta }^{\circ }1$
and $\angle O_{2}AP=\angle O_{2}PA={\theta }^{\circ }2$
then in $\Delta APQ--$
$\because$ $\angle APQ+\angle AQP+\angle PAQ={180}^{\circ }$
$\Rightarrow$ $(90-{\theta }_1^{\circ })+({90}^{\circ }-{\theta }_2^{\circ })+a^{\circ }={180}^{\circ }$
$\therefore$ a = ${\theta }_1+{\theta }_2$ ...(1)
$\because$ The straight line $O_{1}A{O}_2$ joining the centres of the circles
$\Rightarrow$ ${\theta }_1^{\circ }+{\theta }_2^{\circ }+a^{\circ }={180}^{\circ }$
$\Rightarrow$ $a^{\circ }+a^{\circ }={180}^{\circ }$
$\therefore \angle PAQ=a^{\circ }$
$\frac{1}{2}\times {180}^{\circ }={90}^{\circ }$