Solution:

Suppose first that the segment $AB$ passes through one of the intersection points of $\mathcal{C}_1$ and $\mathcal{C}_2$, and let $C$ be that point. Let also $O$, $O_1$, $O_2$ be respectively the centers of the circles $\mathcal{C}$, $\mathcal{C}_1$, $\mathcal{C}_2$. The triangles $OAB$, $O_1AC$ and $O_2BC$ are isosceles and are similar (for example $OAB$ and $O_1AC$ share a base angle). Therefore the opposite sides of the quadrilateral $OO_1CO_2$ are parallel and this quadrilateral is a parallelogram. Hence $r = OA = OO_1 + O_1A = O_2C + O_1A = r_2 + r_1$.

Conversely, suppose $r = r_1 + r_2$. Construct the parallelogram $OO_1C' O_2'$ whose vertices $C'$ and $O_2'$ lie on the segments $AB$ and $OB$ (it is clear that this can be done in a unique way). From the similarity of $O_1AC'$, $O_2'C'B$ and $OAB$ we get that $O_1C' = O_1A = r_1$, from which it follows that $C'$ lies on the circle $\mathcal{C}_1$, and also $O_2'B = O_2'C = OO_1 = OA - O_1A = r - r_1 = r_2$, from which it follows that $O_2' = O_2$ and that $C'$ lies on the circle $\mathcal{C}_2$.