Touchy circles

width=50mm,scale=1.5

width=50mm,scale=1.5

$\therefore$ $\large\dfrac{1}{\sqrt{r_{3}}} = 1 + \dfrac{1}{2}$

$\large r_{3} = \boxed{\dfrac{4}{9}}$

For these type of problems we have direct formula

$\large \dfrac{1}{\sqrt{r_{middle}}} = \dfrac{1}{\sqrt{r_{left}}} + \dfrac{1}{\sqrt{r_{right}}}$

Here unknown is $\large r_{middle}$

$\therefore$ $\large\dfrac{1}{\sqrt{r_{middle}}} = 1 + \dfrac{1}{2}$

$\large r_{middle} = \boxed{\dfrac{4}{9}}$

$\sqrt{(1+r)^2-(1-r)^2} + \sqrt{(4+r)^2-(4-r)^2} = (\sqrt{4+1)^2 - (4-1)^2}$

$\sqrt{(1+2r+r^2)-(1-2r+r^2)} + \sqrt{(16+8r+r^2)-(16-8r+r^2)} = \sqrt{5^2 - 3^2}$

$\sqrt{4r} + \sqrt{16r} = \sqrt{25-9}$

$2\sqrt{r} + 4\sqrt{r} = \sqrt{16}$

$6\sqrt{r} = 4$

$\sqrt{r} = \frac{4}{6} = \frac{2}{3}$

$r = (\frac{2}{3})^2 = \frac{4}{9}$ .

So $a+b = 4+9 = \boxed{13}$ .