Question

Three equal circles each of radius $r$ touch one another.The radius of the circle touching all the three given circles internally is

• A. $(2+\sqrt{3})r$
• B. $\frac{(2+\sqrt{3})}{\sqrt{3}}r$
• C. $\frac{(2-\sqrt{3})}{\sqrt{3}}r$
• D. $(2-\sqrt{3})r$

Answer 1

The correct option is B $\frac{(2+\sqrt{3})}{\sqrt{3}}r$

$△DEF$ is equilateral with side $2r$
If radius of circumcircle $DEF$ is $R_1$
Then,area of $△DEF=\frac{\sqrt{3}}{4}{\left(2r\right)}^2=\sqrt{3}{r}^2$
$\Rightarrow \sqrt{3}{r}^2=\frac{2r.2r.2r}{4R_1}$
$\Rightarrow R_1=\frac{2r}{\sqrt{3}}$
$\therefore$ Radius of the circle touching all the three given circles
$=r+R_1$
$=r+\frac{2r}{\sqrt{3}}$
$=\frac{(2+\sqrt{3})r}{\sqrt{3}}$