A Circle Problem!

$\overline{AQ} = \overline{AV} = 2R - x$ and $\triangle{APS} \sim \triangle{ATO} \implies$ $\dfrac{R - 2r}{r + x} = \dfrac{1}{2}$

$\implies 2R - 4r = r + x \implies x = 2R - 5r \implies \overline{PS} = r + x = 2(R - 2r)$ and

$\overline{AP} = \overline{AV} + r = 6r$

In $\triangle{APS} \implies 4R^2 = 4(R^2 - 4rR + 4r^2) + 36r^2 \implies 16rR = 52r^2 \implies$

$\dfrac{R}{r} = \dfrac{13}{4} = \dfrac{a}{b} \implies a + b = \boxed{17}$ .

From segment $OW$ , $OT = OW - TW = R - 2r$ .

Since $\triangle ATO \sim \triangle APS$ by AA similarity, $PS = OT \cdot \frac{AS}{AO} = (R - 2r) \cdot \frac{2R}{R} = 2R - 4r$ .

By the Pythagorean Theorem on $\triangle APS$ , $AP = \sqrt{AS^2 - PS^2} = \sqrt{(2R)^2 - (2R - 4r)^2} = 2\sqrt{4rR - 4r^2}$ .

As an inradius of a right $\triangle APS$ , $r = \frac{1}{2}(PS + AP - AS) = \frac{1}{2}(2R - 4r + 2\sqrt{4rR - 4r^2} - 2R)$ .

$r = \frac{1}{2}(2R - 4r + 2\sqrt{4rR - 4r^2} - 2R)$ rearranges to $3r = \sqrt{4rR - 4r^2}$ , then to $13r^2 = 4rR$ , and finally to $\frac{R}{r} = \frac{13}{4}$ .

Therefore, $a = 13$ , $b = 4$ , and $a + b = \boxed{17}$ .