Question

Two circle touch each other externally at $C$ and $AB$ is a common tangent to the circles. Then, $\angle ACB=$

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

Answer 1

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

Let $A$ be on the circle with centre $O$ and $B$ be the point on the circle with $O^{\prime }$ as centre and $AB$ be the tangent to both circles touching at $A$ and $B$.

Let the two circles touch at $C$.

Let the tangent at $C$ meet $AB$ at $N$.

Now, $NA$ and $NT$ are tangents to the circle with centre $O$ and therefore $NA=NB$.

So, the $△NAC$ is isosceles and angles $NAC=NCA=x$ ....(say)

By similar way, we know

$NB$ and $NT$ are tangents from $N$ to the circle with centre $O^{\prime }$.

So, $△NBC$ is isosceles with $NB=NC$ and therefore

angles $NBC=NCB=y$ ....(say)

$\therefore$ In $△ABC$, we have

angles $A+B+C=x+y+(x+y)={180}^0$

$2(x+y)={180}^0$

$\Rightarrow x+y=\frac{180}{2}$

$={90}^0$