Explanation

(b)
Since a radius of circle $C_{1}$ is 2 and this circle touches both the axes
So, centre of circle $C_{1} = (2,\ 2)$ and let radius of another circle is $r$ and this circle also touches both the axes so centre of circle $C_{2} = (r,\ r)$
Since, both circles touches each other
$\sqrt{(r - 2)^{2} + (r - 2)^{2}} = 2 + r$
$\Rightarrow \ \ 2(r - 2)^{2} = (r + 2)^{2}$
$\Rightarrow \ \ r^{2} - 12r + 4 = 0$
$\Rightarrow \ \ \ r = \frac{12 \pm \sqrt{128}}{2} = 6 \pm 4\sqrt{2}$
$\Rightarrow \ \ r = 6 + 4\sqrt{2}\ \ \ \ \ \lbrack\because r > 2\rbrack$