Question

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

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

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

Let $\angle CAP=xand\angle CBP=y$
CA = CP [Lengths of the tangents from an external point C]
Therefore in traingle PAC, $\angle CAP=\angle APC=x$
Similarly $CB=CPand\angle CPB=\angle PBC=y$
Now in the triangle APB,3
$\angle PAB+\angle PBA+\angle APB=180$ [sum of the interior angles in a triangle]
$x+y+(x+y)=180$
$2x+2y=180$
$x+y=90$
therefore $\angle APB=x+y=90$