Range for radius of tangent sphere to three given spheres

Question

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 $a = 15, b = 20, c = 25$. 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 $a = 15, b = 20, c = 25$, the three spheres are shown in this 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.

Answer 1

This answer uses $C_1,C_2,C_3$ that you wrote.

You already got $$C_1(0,0,a), C_2 (2\sqrt{ab}, 0, b),C_3(u,v,c)$$ where $$\begin{align}u&=\dfrac{ a b + a c - bc }{ \sqrt{a b}} \\\\v&=\dfrac{\sqrt{2 a b c (a + b + c) - a^2 b^2-b^2c^2 -c^2a^2} }{\sqrt{ab}}\end{align}$$

Let $C_1'(0,0,0),C_2'(2\sqrt{ab},0,0),C_3'(u,v,0)$.

Then, using real numbers $s,t,f$ satisfying $s\ge 0,t\ge 0,s+t\le 1$ and $f\ge 0$, we can write $$C_4(2\sqrt{ab}\ s+ut,vt,f)$$ where $C_4'(2\sqrt{ab}\ s+ut,vt,0)$ is inside $\triangle{C_1'C_2'C_3'}$.

So, letting $d$ be the radius of the fourth sphere, our problem is reduced to finding $d$ such that there are real numbers $s,t,f$ satisfying $$0\le s,0\le t,s+t\le 1,f\ge 0\tag1$$ $$(a+d)^2=(2s\sqrt{ab}+tu)^2+(tv)^2+(f-a)^2\tag2$$ $$(b+d)^2=(2s\sqrt{ab}+tu-2\sqrt{ab})^2+(tv)^2+(f-b)^2\tag3$$ $$(c+d)^2=(2s\sqrt{ab}+tu-u)^2+(tv-v)^2+(f-c)^2\tag4$$

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Let us consider the case where $a = 15, b = 20, c = 25$.

We have $u=\frac{35}{6}\sqrt 3$ and $v=\frac{5}{6}\sqrt{2013}$.

$(2)$ is equivalent to

$$f^2=d^2-1200s^2-700st-1500t^2 + 30 d + 30 f\tag5$$

From $(3)(5)$, we have

$$f=-d - 240 s - 70 t + 120\tag6$$

From $(4)(6)$, we have $$d(24s+7t-12) = -2940 s^2 + 2315s - 1715 st - 320 t^2 + 815 t - 495$$ Suppose here that $24s+7t-12=0$. Then, we have $(t - 1)^2 = -\frac{25}{671}$ which is impossible.

So, we have $$d =\frac{-2940 s^2 + 2315s - 1715 st - 320 t^2 + 815 t - 495}{24s+7t-12}\tag7$$

From $(5)(6)$, we have $$d =\frac{-2940 s^2 + 2520s - 1715 st- 320 t^2 + 735 t - 540}{24s+7t-12}\tag8$$

From $(7)(8)$, we have $$s=\frac{16t+9}{41}\tag9$$

From $(7)(9)$, we have $$d=-\frac{300 (8052 t^2 - 4697 t + 720)}{41 (671 t - 276)}\tag{10}$$

From $(6)(9)(10)$, we have $$f=-\frac{10 (208681 t^2 - 229482 t + 54576)}{41 (671 t - 276)}\tag{11}$$

From $(1)$ with $d\gt 0$, we get $$0\le t\le \frac{114741 - 615\sqrt{4697}}{208681}\approx 0.348$$

From $(10)$, we have $$d'(t)=\frac{-2415600 (671 t^2 - 552 t + 101)}{41 (671 t - 276)^2}$$ from which we get $$d'(t)=0\iff t = \frac{276- 41 \sqrt 5}{671}(\approx 0.275),t=\frac{276+41 \sqrt 5}{671}(\approx 0.548)$$

So, we get $$\min d=d\bigg(\frac{276-41\sqrt 5}{671}\bigg)=\frac{300 (24 \sqrt 5 - 47)}{671}$$ and $$\begin{align}\max d&=\max\bigg(d(0),d\bigg(\frac{114741 - 615 \sqrt{4697}}{208681}\bigg)\bigg) \\\\&=\max\bigg(\frac{18000}{943},\frac{150 (\sqrt{4697} - 47)}{311}\bigg) \\\\&=\frac{18000}{943}\end{align}$$

Therefore, we finally see that the range of $d$ is $$\color{red}{\frac{300 (24 \sqrt 5 - 47)}{671}\le d\le \frac{18000}{943}}$$

Answer 2

The locus of the center $P=(x,y,z)$ of a sphere tangent to spheres $C_1$ and $C_2$ is a sheet of a hyperboloid, whose equation can be found from $PC_2-PC_1=b-a$. Using the formulas for $C_1$, $C_2$, $C_3$ given in the question, this equation can be expanded to: $$ 4 (a-b)^2 \left(\left(x-2 \sqrt{a b}\right)^2+(b-z)^2+y^2\right)=4 \left(-2 x \sqrt{a b}+a (b+z)+b^2-b z\right)^2. $$ Analogously, the locus of the center $P=(x,y,z)$ of a sphere tangent to spheres $C_1$ and $C_3$ is a sheet of another hyperboloid, with equation: $$ 4 (a-c)^2 \left(\left(\frac{b c-a (b+c)}{\sqrt{a b}}+x\right)^2+\left(\sqrt{4 a c-\frac{(b c-a (b+c))^2}{a b}}-y\right)^2+(c-z)^2\right)=4 \left(-c \left(\frac{a x}{\sqrt{a b}}+z\right)+\frac{b c x}{\sqrt{a b}}-y \sqrt{4 a c-\frac{(b c-a (b+c))^2}{a b}}-x \sqrt{a b}+a (c+z)+c^2\right)^2. $$ The intersection of those two hyperboloids gives then the locus of the center of the fourth sphere, with center $P$ and tangent to the other three spheres. The limit position of $P$, to avoid the fourth sphere to fall, is on the plane $y=0$, which is the plane passing through the centers of the two smallest spheres and perpendicular to $(xy)$ plane. Hence this position of $P$ can be found by solving the system formed by the two above equations and $y=0$.

This can also be seen as the intersection of two hyperbolas (shown in figure below), whose equations in the $(xz)$ plane are obtained by plugging $y=0$ in the above equations.

Unfortunately, the resultant equation is a cubic, leading to three solutions quite long and involved to write down in their general form, as a function of $a$, $b$ and $c$. For $𝑎=15$, $𝑏=20$, $𝑐=25$ I got (with the help of Mathematica): $$ \begin{align} P_1&=\left({180\sqrt3\over41},0,{45480\over943}\right)\\ P_2&=\left(\frac{5}{4} \sqrt{\frac{3}{2} \left(889+37 \sqrt{409}\right)} ,0,-\frac{25}{64} \left(119+11 \sqrt{409}\right)\right)\\ P_3&=\left(\frac{5}{4} \sqrt{\frac{3}{2} \left(889-37 \sqrt{409}\right)} ,0,\frac{25}{64} \left(-119+11 \sqrt{409}\right)\right)\\ \end{align} $$ But only the first solution corresponds to a point lying on the "right" sheet of the hyperboloids, i.e. $P_1$ is the only position of $P$ satisfying $PC_1-a=PC_2-b=PC_3-c=r$. The corresponding radius is: $$ r=\frac{18000}{943}. $$ This is then the maximum radius of the fourth sphere.

Answer 3

Comment: I could find the radius of fourth circle using CAD software:

$r\approx 7.585$ and $r\approx 24.765$ for $a=75$, $b=50$, $c=35$.

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