Question

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

Answer: (B), (C)

$\frac{(2-\sqrt{3})}{\sqrt{3}}r$
$\frac{(2+\sqrt{3})}{\sqrt{3}}r$

$△DEF$ is equilateral with side $2r$

Let the radius of circumcircle of triangle $DEF$ be $R_1$.

Now, area of $△DEF=\frac{\sqrt{3}}{4}{\left(2r\right)}^2=\sqrt{3}{r}^2$

$\Rightarrow \sqrt{3}{r}^2=\frac{2r.2r.2r}{4R_1}\Rightarrow R_1=\frac{2r}{\sqrt{3}}(\because \Delta =\frac{abc}{4R})$

radius of the circle touching all the three given circles

$=R_1\pm r\frac{(2\pm \sqrt{3})}{\sqrt{3}}r$