Question

In the given figure, two circles touch each other externally at point $P$. $AB$ is the direct common tangent of these circles. Prove that :

(i) tangent at point $P$ bisects $AB$,

(ii) angle $APB=$ ${90}^{\circ }$.

Answer 1

given: $C_1$ and $C_2$ are two circles that touch each other externally at P. AB is the common tangent to the circles $C_1$ and $C_2$ at points $A$and $B$ respectively.

TPT: $\angle APB={90}^{\circ }$.

Proof: let $\angle CAP=x$ and $\angle CBP=y$

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

therefore in triangle $PAC$, $\angle CAP=\angle APC=x$

similarly $CB=CP$ and $\angle CPB=\angle PBC=y$

Now in the triangle $APB$,

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

$x+y+(x+y)={180}^{\circ }$

$2x+2y={180}^{\circ }$

$x+y={90}^{\circ }$

therefore $\angle APB=x+y={90}^{\circ }$

Hence $AB$ subtends a right angle at $P$ and $\angle APB={90}^{\circ }$