Question

Two circles touch each other externally at P. AB is a 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 ${90}^{\circ }$

Given X and Y are two circles touch each other externally at P. AB is the common tangent to the circles X and Y at point A and B respectively.

let $\angle CAP=\alpha$ and $\angle CBP=\beta$.

CA = CP [lengths of the tangents from an external point C].

In a triangle PAC, $\angle CAP=\angle APC=\alpha$

Similarly CB = CP and $\angle CPB=\angle PBC=\beta$

Now in the triangle APB,

$\angle PAB+\angle PBA+\angle APB={180}^o$ [sum of the interior angles in a triangle]

$\alpha +\beta +(\alpha +\beta )={180}^o$

$2\alpha +2\beta ={180}^o$

$\alpha +\beta ={90}^o$

Thereofre, $\angle APB=\alpha +\beta ={90}^o$