Question

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

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

Since, $AB=BC=CA=2r$

Therefore, $ABC$ is an equilateral triangle

Area of the triangle $ABC=\frac{1}{4}\sqrt{3}\times {\left(2r\right)}^2=\sqrt{3}{r}^2$

$\therefore$ Area of $△AOB=\frac{1}{3}\times$ Area of of $△ABC=\frac{1}{3}\times \sqrt{3}{r}^2=\frac{r^2}{\sqrt{3}}$

Also area of $△AOB=\frac{1}{2}\left(2r\right)\times y=ry$ ( Let $ON=y$)

$\therefore ry=\frac{r^2}{\sqrt{3}}\Rightarrow y=\frac{r}{\sqrt{3}}$

$\therefore$ In $△AON$

${(r+x)}^2=r^2+y^2=r^2+{\left(\frac{r}{\sqrt{3}}\right)}^2=\frac{4r^2}{3}\Rightarrow x+y=\frac{2r}{\sqrt{3}}$

$\therefore x=(\frac{2}{\sqrt{3}}-1)r=\left(\frac{2-\sqrt{3}}{\sqrt{3}}\right)r$