The key here is to realize that if a right triangle is inscribed in a circle then the hypotenuse is the diameter of the circle. Thus $2r_1$ is the length of $AC$ and $2r_2$ is the length of $BC.$ By Pythagorean Theorem $AB=\sqrt{(2r_1)^2+(2r_2)^2}=2\sqrt{r_1^2+r_2^2}.$ Since the radius of an inscirbed circle of a right triangle with legs $a,b$ and hypotenuse $c$ is $\frac{a+b-c}{2}.$ We can apply that here. Thus our answer is $\frac{2r_1+2r_2-2\sqrt{r_1^2+r_2^2}}{2}=\boxed{r_1+r_2-\sqrt{r_1^2+r_2^2}}$