The largest circle

$\frac{r}{OC}=sin(30^\circ)=\frac{1}{2}$ therefore $OC=2\times r$

Since $OC+r=1$ we have $3\times r=1$ or $r=\frac{1}{3}$

We desire a circle whose center is on the angle bisector of $\angle ABC$ and lies tangent to the two radii that create the sector and the larger circle itself. As such, we can draw lines through the center of the smaller circle from the points of tangency to the tangent that passes through point D. Extending the two radii to these points, we obtain $\triangle ABC$ . Notice that $\triangle ABC$ is equilateral as $\angle ADC$ and $\angle ADB$ are right and $\angle CAD$ and $\angle BAD$ are $30^{\circ}$ in measure. Thus the circle we desire is the incircle of the equilateral $\triangle ABC$ .

To find the side length use the Pythagorean theorem where $AD$ measures 1, $DC$ measures $\frac{S}{2}$ , and $AC$ measures $S$ . this yields $1+\frac{S^2}{4}=S^2$ . Simplifying and gathering like-terms we get $\frac{4}{3}=S^2$ or $S=\frac{2}{\sqrt{3}}$ . As the radius of the incircle of an equilateral triangle is $\frac{S}{2\sqrt{3}}$ , the radius of the circle is $\frac{1}{3}$ . It follows then that $a+b=4$ .