As shown in figure, three circles which  have the same radius $r$ , have centres at  $(0,0),(1,1)$ and $(2,1)$ . If they have a common tangent line, as shown, then the value of  $10\sqrt{5}r$ is 

Equation of line joining origin and centre of circle 
$C_2\equiv (2,1)$  is,  $y=\frac{x}{2}$
 $\Rightarrow x-2y=0$
Let equation of common tangent is  $x-2y+c=0\left(1\right)$
 perpendicular distance from $(0,0)$  on this line 
 perpendicular distance from  $(1,1)$
 $$\begin{gathered} \Rightarrow \left|\frac{c}{\sqrt{5}}\right|=\left|\frac{c-1}{\sqrt{5}}\right| \\ \Rightarrow c=1-c\Rightarrow c=\frac{1}{2} \end{gathered}$$
 
Equation of common tangent is 
 $x-2y+\frac{1}{2}=0or2x-4y+1=0\left(2\right)$
Perpendicular from  $(2,1)$ on the line  $\left(2\right)$
 $r=\left|\frac{4-4+1}{\sqrt{20}}\right|=\frac{1}{2\sqrt{5}}=\frac{\sqrt{5}}{10}$