Mysterious Radii

If we draw the radii OA & QB, by isosceles angles, $\angle OAP = \angle APO$ and $\angle QPB = \angle QBP$ , and $\angle APO =\angle BPQ$ due to intersection angles. Thus, all four angles are equal and measured as $x$ , as colored as yellow in the image above.

Then we can deduce that the triangles OPA and QPB are similar, resulting in equal ratios: $AP:PB = OP:PQ$ .

Again with intersection angles $\angle OPQ = \angle APB$ , we can conclude with the same side ratios that the triangles OPQ & APB are also similar, and $x = \angle PAB + \angle PBA$ .

Finally, by calculating all four angles around the ABQO, the angles are supplementary, summing up to 180°, making the quadrilateral ABQO cyclic.

Hence, by Ptolemy's theorem $8\times 7 = 4\times 8 + Rr$ . $Rr =24$ .

Since the big radius can not exceed OQ=8. Only R=6 and r =4 apply. As a result, $R+r = \boxed{10}$ .