TITLE: Proof of Descartes' theorem
QUESTION [5 upvotes]: I came across the use of Descartes' theorem while solving a question.I searched it but I  could only find the theorem but not any proof.Even Wikipedia also, just states the theorem!!I want to know the procedure to find the radius of the Soddy Circle??
I apologize if its duplicate and to mention it is not a homework.

REPLY [11 votes]: In high-school, my geometry teacher mentioned the Soddy-Gosset theorem and I had been curious about its proof ever since. About 5 years ago, I decided to attempt to prove it and came up with a nice proof and a corollary that, according to my investigation, was unknown, at least in the form that I have.
The first section below is taken from a paper I wrote at that time. The later sections are a clarification of the later parts of that paper.

Tangent Spheres
Suppose that the sphere with radius $r_1$ and center $c_1$, $(c_1,r_1)$, and the sphere with radius $r_2$ and center $c_2$, $(c_2,r_2)$, are externally tangent. Then we have
$$
\begin{align}
|c_1-c_2|^2&=(r_1+r_2)^2\tag{1a}\\
|c_1|^2-2c_1\cdot c_2+|c_2|^2&=r_1^2+2r_1r_2+r_2^2\tag{1b}
\end{align}
$$
For a given sphere, $(c,r)$, the radius of the inverted sphere, $\bar{r}$, satisfies
$$
\begin{align}
\frac{|c|^2}{r^2}-1&=\frac1{r\bar{r}}\tag{2a}\\
|c|^2-r^2&=\frac r{\bar{r}}\tag{2b}
\end{align}
$$
We don't actually use any of the properties of spherical inversion, so we can simply take $(2)$ as a definition of $\bar{r}$.
Thus, if we subtract $r_1^2+r_2^2$ from both sides of $\mathrm{(1b)}$ and use equation $\mathrm{(2b)}$, we get
$$
\begin{align}
\frac{r_1}{\bar{r}_1}+\frac{r_2}{\bar{r}_2}-2c_1\cdot c_2&=2r_1r_2\tag{3a}\\
\frac{c_1}{r_1}\cdot\frac{c_2}{r_2}-\frac{1}{2}(\frac{1}{\bar{r}_1r_2}+\frac{1}{r_1\bar{r}_2})&=-1\tag{3b}
\end{align}
$$
Furthermore, from equation $\mathrm{(2a)}$, we get
$$
\frac{|c|^2}{r^2}-\frac{1}{r\overline{r}}=1\tag{4}
$$
Therefore, if, for a sphere in $\mathbb{R}^n$, $(c,r)$, we define the row vector in $\mathbb{R}^{n+2}$
$$
v=\left\langle\frac cr,\frac1r,\frac1{\bar{r}}\right\rangle\tag{5}
$$
and make a square matrix, $V$, from $n+2$ of these row vectors for $n+2$ mutually tangent spheres, then we get from equation $\mathrm{(3b)}$, for the off-diagonal elements, and equation $(4)$, for the diagonal elements, that
$$
V\left[\begin{array}{c c c}
I_n & 0_{n\times 1} & 0_{n\times 1} \\
0_{1\times n} & 0 & -\frac{1}{2} \\
0_{1\times n} & -\frac{1}{2} & 0
\end{array}\right]V^T = 2I_{n+2}-C_{n+2}\tag{6}
$$
where $C_{n+2}$ is an $(n{+}2)\times(n{+}2)$ matrix of all $1$s. Because $C_{n+2}^2=(n+2)C_{n+2}$, it is easily verified that
$$
(2I_{n+2}-C_{n+2})^{-1}=\frac{1}{2}I_{n+2}-\frac{1}{2n}C_{n+2}\tag{7}
$$
Invert equation $(6)$ and move $V$ to the right side to get
$$
\left[\begin{array}{c c c}
I_n & 0_{n\times 1} & 0_{n\times 1} \\
0_{1\times n} & 0 & -2 \\
0_{1\times n} & -2 & 0
\end{array}\right] = V^T(\frac{1}{2}I_{n+2}-\frac{1}{2n}C_{n+2})V\tag{8}
$$
Looking at the $(n+1,n+1)$ element of the matrix equation $(8)$, we get the equation
$$
n\sum_{k=1}^{n+2}\frac{1}{r_k^2}=\left(\sum_{k=1}^{n+2}\frac{1}{r_k}\right)^2\tag{9}
$$
If one of the spheres encompasses the rest, so that all the other spheres are internally tangent to it, then equation $\mathrm{(1a)}$ becomes $|c_1-c_2|^2=(r_1-r_2)^2$. If we consider the radius of the encompassing sphere to be negative, then this becomes $|c_1-c_2|^2=(r_1+r_2)^2$ as before. Therefore, we will consider the radius of an encompassing sphere to be negative.
Thus, we have shown

The Soddy-Gosset Theorem: Given $n+2$ mutually tangent spheres in $\mathbb{R}^n$, their radii, $\{r_k:1\le k\le n+2\}$ satisfy equation $(9)$ if the radius of an encompassing sphere is considered negative.

Furthermore, if we look at the $(j,n+1)$ elements of the matrix equation $(8)$, for $1\le j\le n$, we get the vector equation
$$
n\sum_{k=1}^{n+2}\frac{c_k}{r_k^2}=\sum_{k=1}^{n+2}\frac{c_k}{r_k}\;\sum_{k=1}^{n+2}\frac{1}{r_k}\tag{10}
$$
If we divide equation $(10)$ by equation $(9)$, we get
$$
\frac{\displaystyle\sum_{k=1}^{n+2}\frac{c_k}{r_k^2}}{\displaystyle\sum_{k=1}^{n+2}\frac{1}{r_k^2}}
=\frac{\displaystyle\sum_{k=1}^{n+2}\frac{c_k}{r_k}}{\displaystyle\sum_{k=1}^{n+2}\frac{1}{r_k}}\tag{11}
$$
Thus, we have derived the following

Corollary: Given $n+2$ mutually tangent spheres in $\mathbb{R}^n$ with radii $\{r_k:1\le k\le n+2\}$, the mean of the centers weighted by $\frac1{r_k^2}$ is equal to the mean of the centers weighted by $\frac1{r_k}$.


Conclusions from Soddy-Gosset
Given the radii of $n+1$ mutually tangent spheres in $\mathbb{R}^n$, the Soddy-Gosset Theorem allows us to compute the radius of two possible spheres that are tangent to the other $n+1$. That is, we can solve $(9)$ for $\frac1{r_{n+2}}$:
$$
n\left(\color{#C00000}{\frac1{r_{n+2}^2}}+\sum_{k=1}^{n+1}\frac1{r_k^2}\right)
=\color{#C00000}{\frac1{r_{n+2}^2}}+2\color{#C00000}{\frac1{r_{n+2}}}\sum_{k=1}^{n+1}\frac1{r_k}+\left(\sum_{k=1}^{n+1}\frac1{r_k}\right)^2\tag{12a}
$$
$$
(n-1)\color{#C00000}{\frac1{r_{n+2}^2}}-2\color{#C00000}{\frac1{r_{n+2}}}\sum_{k=1}^{n+1}\frac1{r_k}+n\sum_{k=1}^{n+1}\frac1{r_k^2}-\left(\sum_{k=1}^{n+1}\frac1{r_k}\right)^2=0\tag{12b}
$$
applying the quadratic formula to $(12)$, we get
$$
\frac{n-1}{r_{n+2}}=\sum\limits_{k=1}^{n+1}\frac1{r_k}\pm\sqrt{n}\left[\left(\sum\limits_{k=1}^{n+1}\frac1{r_k}\right)^2-(n-1)\sum\limits_{k=1}^{n+1}\frac1{r_k^2}\right]^{1/2}\tag{13}
$$
Note that the discriminant in $(13)$ vanishes precisely when the first $n+1$ spheres satisfy the Soddy-Gosset Theorem on $\mathbb{R}^{n-1}$. This implies that $\{c_k:1\le k\le n+1\}$ lie in an $n-1$ dimensional hyperplane, and the two possibilities for $(c_{n+2},r_{n+2})$ are reflections of each other across this hyperplane.

Conclusions from the Corollary
Given the radii of $n+2$ mutually tangent spheres and the centers of the first $n+1$ of them, the Corollary allows us to determine $c_{n+2}$, except in the case where the coefficients of $c_{n+2}$ cancel in $(11)$. That is, when
$$
\frac{\dfrac1{r_{n+2}^2}}{\displaystyle\sum_{k=1}^{n+2}\frac{1}{r_k^2}}
=\frac{\dfrac1{r_{n+2}}}{\displaystyle\sum_{k=1}^{n+2}\frac{1}{r_k}}\tag{14}
$$
which is equivalent to
$$
\begin{align}
\frac1{r_{n+2}}
&=\frac{\displaystyle\sum_{k=1}^{n+2}\frac1{r_k^2}}{\displaystyle\sum_{k=1}^{n+2}\frac1{r_k}}
=\frac{\displaystyle\sum_{k=1}^{n+1}\frac1{r_k^2}}{\displaystyle\sum_{k=1}^{n+1}\frac1{r_k}}\tag{15a,15b}\\
&=\frac1n\sum_{k=1}^{n+2}\frac1{r_k}
=\frac1{n-1}\sum_{k=1}^{n+1}\frac1{r_k}\tag{15c,15d}
\end{align}
$$
Explanation:
$\mathrm{(15a)}$: restatement of $(14)$
$\mathrm{(15b)}$: multiply $\mathrm{(15a)}$ by $\sum\limits_{k=1}^{n+2}\frac1{r_k}$, subtract $\frac1{r_{n+2}^2}$, then divide by $\sum\limits_{k=1}^{n+1}\frac1{r_k}$
$\mathrm{(15c)}$: apply $(9)$ to $\mathrm{(15a)}$
$\mathrm{(15d)}$: multiply $\mathrm{(15c)}$ by $n$, subtract $\frac1{r_{n+2}}$, then divide by $n-1$
Note that $\mathrm{(15b)}$ and $\mathrm{(15d)}$ say that the first $n+1$ spheres satisfy the Soddy-Gosset Theorem in $\mathbb{R}^{n-1}$. Again, this implies that $\{c_k:1\le k\le n+1\}$ lie in an $n-1$ dimensional hyperplane.
Furthermore, $\mathrm{(15d)}$ matches $(13)$ when the discriminant vanishes.

Locating $\boldsymbol{c_{n+2}}$ in the Exceptional Case
By rotation and translation, we can arrange that the $n-1$ dimensional hyperplane containing $\{c_k:1\le k\le n+1\}$ is $x_n=0$. That is, if $1\le k\le n+1$,
$$
c_{k,n}=0\tag{16}
$$
and $c_{n+2,n}$ is the perpendicular distance of $c_{n+2}$ above or below that hyperplane. For clarity of presentation, let $\hat{p}$ be the projection of $p$ onto the hyperplane $x_n=0$; that is, $\hat{p}_k=p_k$ for $1\le k\le n-1$ and $\hat{p}_n=0$.
Looking at the $(j,n)$ elements of matrix equation $(8)$ for $1\le j\le n-1$, we get the vector equation
$$
\begin{align}
\frac{\hat{c}_{n+2}}{r_{n+2}}
&=\frac1n\sum_{k=1}^{n+2}\frac{\hat{c}_k}{r_k}\tag{17a}\\
&=\frac1{n-1}\sum_{k=1}^{n+1}\frac{c_k}{r_k}\tag{17b}
\end{align}
$$
which, in light of $\mathrm{(15d)}$, gives the vector equation
$$
\hat{c}_{n+2}=\frac{\displaystyle\sum_{k=1}^{n+1}\frac{c_k}{r_k}}{\displaystyle\sum_{k=1}^{n+1}\frac1{r_k}}\tag{18}
$$
Looking at the $(n,n)$ element of matrix equation $(8)$ we get
$$
\begin{align}
1
&=\frac12\frac{c_{n+2,n}^2}{r_{n+2}^2}-\frac1{2n}\frac{c_{n+2,n}^2}{r_{n+2}^2}\\
&=\frac{n-1}{2n}\frac{c_{n+2,n}^2}{r_{n+2}^2}\tag{19}
\end{align}
$$
which gives us
$$
c_{n+2,n}=\pm r_{n+2}\sqrt{\frac{2n}{n-1}}\tag{20}
$$
We can now remove the rotation, and use $(18)$ and $(20)$ to get that
$$
c_{n+2}=\frac{\displaystyle\sum_{k=1}^{n+1}\frac{c_k}{r_k}}{\displaystyle\sum_{k=1}^{n+1}\frac1{r_k}}\pm ur_{n+2}\sqrt{\frac{2n}{n-1}}\tag{21}
$$
or equivalently, using Soddy-Gosset and the Corollary in $\mathbb{R}^{n-1}$ and $\mathrm{(15b)}$,
$$
\frac{c_{n+2}}{r_{n+2}}=\frac{\displaystyle\sum_{k=1}^{n+1}\frac{c_k}{r_k}\frac1{r_k}}{\displaystyle\sum_{k=1}^{n+1}\frac1{r_k}}\pm u\sqrt{\frac{2n}{n-1}}\tag{22}
$$
where $u$ is a unit vector perpendicular to the plane containing $\{c_k:1\le k\le n+1\}$.