There are four circles having radii $r_1, r_2, r_3 $ and $r_4$ touching one another on a spherical surface of radius $R$ (as shown in the picture below, four colored circles touching one another at 6 points on the sphere). Any help to find out the radius $R$ in terms of $r_1, r_2, r_3 $ and $r_4$?
Assume that $r_1, r_2, r_3, r_4<R$
On
(Since you didn't specify, I'm assuming that the distances $r_1, ...$ are distances on the surface, not in $\mathbb{R}^3$. Sorry if this turns out to be useless!)
Draw the lines between the centers of the circles. This divides the sphere into four spherical triangles -- making it a "spherical tetrahedron", if you will. We know the side lengths of the triangles: the sum of the radii of the circles. For instance, one triangle has side lengths $\{r_1+r_2, r_1+r_3, r_2+r_3\}$.
Then we can find the area $\Delta$ of each triangle. To do so we find the angles using the spherical law of cosines, and then the spherical triangle area formula (see this answer).
$$a = \frac{r_1+r_2}{R}$$ $$C = \arccos\left(\frac{\cos c-\cos a \cos b}{\sin a \sin b}\right)$$ $$\Delta = R^2 (A+B+C-\pi)$$ By substituting these values in for the four sides, twelve angles, and four triangles, and requiring that their areas sum to $4\pi R^2$, we can build an equation for $R$. This can provide the value of having an equation in only one variable to solve -- but I worry that it will be far from solvable by anything other than numerical methods.
Sorry, this is not a complete answer, but I don't have enough reputation to simply comment. As a starting point I think you need some constraints, for example, it appears from the image that each circle touches all three other circles - I will assume that this a requirement. Also, as long as it is not necessary that $r1,r2,r3,r4 < R$, then perhaps it would simplify things to consider the special case of $r1 = R$, and then $r2,r3,r4$ reside in a hemisphere (assuming that all circles must touch three other circles).