You're given three spheres of radii $a, b, c$ where $a \le b \le c $. The three spheres are placed on a table (represented by the $xy$ plane), such that they tangent to each other. I want to find the range for the radius of a fourth sphere that is tangent to the three spheres from the outside, such that the fourth sphere does not "fall off".
As a specific example, let $a = 7, b = 10, c = 15$$a = 15, b = 20, c = 25$. What would the range for radius of fourth sphere be in this case?
My attempt:
Let the first sphere (the smallest one) by located at $ C_1 = (0,0, a)$, and let the second sphere be located at $C_2 = (x, 0, b)$, where $ x \gt 0$, then necessarily
$(a + b)^2 = x^2 + (a - b)^2 $
Therefore, $ x = 2 \sqrt{a b} $
Next, take the third sphere and place it at $C_3 = (u, v, c)$, where $ v \gt 0 $, then
$ (a + c)^2 = u^2 + v^2 + (c - a)^2 $
$ (b + c)^2 = (u - x)^2 + v^2 + (c - b)^2 $
These two equations reduce to
$ u^2 + v^2 = 4 a c $
$ u^2 + v^2 - 2 x u + x^2 = 4 b c $
Subtracting, gives us $u$ as
$ u = \dfrac{ a b + a c - bc }{ \sqrt{a b}} $
Then $v$ follows:
$ v = \sqrt{ 4 a c - u^2 } = \dfrac{\sqrt{2 a b c (a + b + c) - a^2 b^2 - a^2 c^2 - b^2 c^2} }{\sqrt{ab} } $
So now the position of the centers of the three balls $C_1, C_2, C_3$ is known.
For $a = 7, b = 10, c = 15$$a = 15, b = 20, c = 25$, the three spheres are shown in this desmos graphicthis desmos graphic
As the for the minimum radius possible for tangency, it is given by the kissing circles theorem of Descartes, namely
$ \dfrac{1}{d} = \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} + 2 \sqrt{ \dfrac{1}{ab} + \dfrac{1}{ac} + \dfrac{1}{bc} } $
where $d$ is the radius of the fourth sphere.
The task remaining, is to find the the upper bound for the radius such that the fourth sphere does not "fall off" the assembly of the first three spheres.