The 4 tennis balls are centered at the vertices of a regular tetrahedron ABCD.

Obviously, AB = BC = AC = AD = BD = CD = 2R

From the top view AH = (2/3) (2R)^{ }√3/2 = 2R/√3

From the side view, AD² = 4R² = AH² + DH²
hence DH² = 4R²(1 - 1/3) = 8R²/3 gives the altitude of the tetrahedron

DH^{ }= 2R√2/√3 = 2/3 R√6

Then from the side view, we "inflate" triangle ADH into triangle FSO,
the side AD translated by R, to FS. And side AH translated by R to FO.

Considering similar triangles to ADH, we finally get the size of triangle FSO, that is
FO = base radius of cone, SO = height of cone.

The figure shows in cyan the construction lines used to send points from side view to top view etc...

In blue are the cone contour, and the touch points T,U,V with the spheres, as well as the touch
circle (T') with sphere (D).

Red circles are the sphere's **contours**, magenta for hidden parts.