Question

As shown in the 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, their radius ${}^{\prime }{r}^{\prime }$ is

• A. $\frac{\sqrt{5}-1}{2}$
• B. $\frac{\sqrt{5}}{10}$
• C. $\frac{1}{2}$
• D. $\frac{\sqrt{3}-1}{2}$

Answer 1

The correct option is A $\frac{\sqrt{5}-1}{2}$

Let the tangent be $y=mx+c$

Distance of this line from $(0,0),(1,1)$ and$2,1)=r$

$\Rightarrow \frac{\left|c\right|}{\sqrt{m^2+1}}=\frac{|m+c-1|}{\sqrt{m^2+1}}=\frac{|2m+c-1|}{\sqrt{m^2+1}}=r$

$\Rightarrow$ Let $\frac{-c}{\sqrt{m^2+1}}=\frac{m+c-1}{\sqrt{m^2+1}}=\frac{-(2m+c-1)}{\sqrt{m^2+1}}=r$

$\Rightarrow \frac{-c}{\sqrt{m^2+1}}=\frac{-2m-c+1}{\sqrt{m^2+1}}\to \left(1\right)$

$\Rightarrow m=\frac{1}{2}$

and $\frac{-c}{\sqrt{m^2+1}}=\frac{m+c-1}{\sqrt{m^2+1}}\to \left(2\right)$

$\Rightarrow c=\frac{1-m}{2}=\frac{1}{4}$

$r=\frac{\left|c\right|}{\sqrt{m^2+1}}$

$=\frac{\frac{1}{4}}{\sqrt{\frac{1}{4}+1}}$

$=\frac{1}{4\sqrt{5}}\sqrt{4}$

$=\frac{2}{4\sqrt{5}}$

$=\frac{2\sqrt{5}}{4\times 5}$

$=\frac{\sqrt{5}}{10}$

$\therefore B)$ Answer.