Four circles touching one another on a spherical surface 
The diagram above shows four identical circles, each having a flat radius $r$ (i.e. flat area $\pi r^2$), touching one another at six different points (i.e. each of four identical circles touches rest three at three different points) on a spherical surface with a radius $R$. How to find out the radius $R$ (of sphere) in terms of radius $r$ (of circles)?
Any help is greatly appreciated.
 A: The six points where a pair of circles touch are the vertices of a regular octahedron, with each face inscribed in one of the circles. Picking these vertices to be the standard basis vectors and their negatives, it's easy to express the radius $R$ of the sphere as a multiple of the radius $r$ of the circles.

A: The six points of intersection of any couple of circles give a octahedron with side length $\sqrt{3}\, r$.
Since the octahedron is inscribed in the sphere,
$$\sqrt{3}\,r = \sqrt{2}\,R$$
follows from the pythagorean theorem.
A: 
Here is a different approach, 

Assume a regular tetrahedron of edge length $a$ whose each face contains each of four equal circles of radius $r$. then the radius of equilateral triangle face 
$$r=\frac{a}{2\sqrt 3}\implies a=2r\sqrt 3$$
The six points of intersection of circles lie on mid-points of six edges of tetrahedron. Now, vertical distance $h$ of center of each circle or face from the center of a regular tetrahedron 
$$h=\frac{a}{2\sqrt 6}$$$$=\frac{2r\sqrt 3}{2\sqrt 6}=\frac{r}{\sqrt 2}$$
join the center of one circle, center of tetrahedron & one of six points of intersection by straight lines that gives a right triangle whose hypotenuse is $R$, and legs $r$ & $h=\frac{r}{\sqrt 2}$, by using pythagorean theorem $$R^2=r^2+\left(\frac{r}{\sqrt 2}\right)^2=\frac32r^2$$  $$R=r\sqrt{\frac{3}{2}}$$
