Question

Let two circles touch each other externally and $P$ is the point of intersection of there direct common tangents. If the direct common tangents from $P$ touches the circles at $A$ and $B$ on same side and $PA=AB=4$, then the radius of smaller circle (in units) is

• A. $4$
• B. $2$
• C. $\sqrt{2}$
• D. $2\sqrt{2}$

Answer 1

The correct option is C $\sqrt{2}$


Let the radius of the small circle be $r$,
As $△PA{C}_1$ and $△PB{C}_2$ are similar, so
$\frac{PA}{PB}=\frac{A{C}_1}{B{C}_2}\Rightarrow \frac{4}{4+4}=\frac{r}{B{C}_2}\Rightarrow B{C}_2=2r$
We know the length of direct common tangent is
$AB=\sqrt{d^2-(r_1-r_2{)}^2}\Rightarrow 4=\sqrt{(3r{)}^2-(2r-r{)}^2}\Rightarrow r=\sqrt{2}$