Circles Part 2

$c_1: x^2+y^2=r^2 ; \text{ center } =(0,0) \text{ and radius } = r$$c_2:(x-1)^2+(y-1)^2=r^2 ; \text{ center } =(1,1) \text{ and radius } = r$$c_3:(x-2)^2+(y-1)^2=r^2 ; \text{ center } =(2,1) \text{ and radius } = r$Distance of $y=m x+c$ line from center $(0,0)$ is,$\left|\frac{0+0+c}{\sqrt{m^2+1}}\right| = r \ldots (1)$Distance of $y=m x+c$ line from center $(1,1)$ is,$\left|\frac{m-1+c}{\sqrt{m^2+1}}\right| = r \ldots (2)$Distance of $y=m x+c$ line from center $(2,1)$ is,$\left|\frac{2m-1+c}{\sqrt{m^2+1}}\right| = r \ldots \text{ (3) }$From (1) and (2), we get$\left|\frac{c}{\sqrt{1+m^2}}\right| = \left|\frac{m-1+c}{\sqrt{1+m^2}}\right|$$\Rightarrow m-1+c = \pm c$taking positive sign,$m-1+c = c$$\Rightarrow m-1 = 0$ From (2) and (3), we get$\left|\frac{m-1+c}{\sqrt{1+m^2}}\right| = \left|\frac{2m-1+c}{\sqrt{m^2+1}}\right|$$\Rightarrow (m-1+c) = \pm (2m-1+c)$taking positive sign,$m-1+c = 2m-1+c$$\Rightarrow m = 0$By taking positive sign we get two different value of $m$ so it is not acceptable.From equation (4), taking negative sign,$m-1+c = -c$$\Rightarrow m + 2c - 1 = 0$From equation (5), taking negative sign$m-1+c = -(2m-1+c)$$\Rightarrow 3m + 2c - 2 = 0 \ldots (7)$Solving equation (6) and (7), we get$3m + 1 - m - 2 = 0$ $\Rightarrow 2m = 1$$\Rightarrow m = \frac{1}{2}$$\therefore 2c = 1 - \frac{1}{2}$$\Rightarrow c = \frac{1}{4}$Putting value of $m = \frac{1}{2}$ and $c = \frac{1}{4}$ in equation (1), we get$r = \left| \frac{\frac{1}{4}}{\sqrt{1 + \frac{1}{4}}} \right|$$= \left| \frac{1}{4} \times \frac{2}{\sqrt{5}} \right|$$= \frac{1}{2\sqrt{5}}$$\therefore 20 (r^2 + c)$$= 20 \left( \frac{1}{4 \times 5} + \frac{1}{4} \right)$$= 20 \left( \frac{1+5}{20} \right)$$= 6$