Jonah Circles
====
# The question
Jonah noticed that given two spheres $S_{r_0}(x_0)$ and $S_{r_1}(x_1)$ given algebraically as $\langle x-x_0,x-x_0\rangle-r_0^2=0$ and $\langle x-x_1,x-x_1\rangle-r_1^2=0$, if the two equations are added then the result $\langle x-x_0,x-x_0\rangle+\langle x-x_1,x-x_1\rangle-(r_0^2+r_1^2)=0$ also results in a sphere. Namely using the bi-linearity and symmetry of the real inner-product
$$
\begin{gather*}
\langle x-x_0,x-x_0\rangle+\langle x-x_1,x-x_1\rangle-(r_0^2+r_1^2)=0\\
\langle x-x_0,x-x_0\rangle+\langle x-x_0,x_0-x_1\rangle-\langle x-x_0,x_0-x_1\rangle+\langle x-x_1,x-x_1\rangle-(r_0^2+r_1^2)=0\\
(\langle x-x_0,x-x_0\rangle+\langle x-x_0,x_0-x_1\rangle)+(\langle x-x_0,x-x_1\rangle-\langle x_1-x_0,x-x_1\rangle)-(r_0^2+r_1^2)=0\\
\langle x-x_0,x-x_1\rangle+\langle x-x_0,x-x_1\rangle-(r_0^2+r_1^2)=0\\
\langle x-x_0,x-x_1\rangle-\frac{r_0^2+r_1^2}{2}=0
\end{gather*}
$$
What is left is to see that the final equation describes a sphere.

The actual question Jonah asked was to descibe what is going on here geometrically, not algebraically. My point was that if you use just the properties of inner-products, then the computation actually shows the geometry.

# Another formula for a sphere
A general form for a sphere can is:
$$\langle x - x_0, x - x_1 \rangle =s$$
with certain conditions on $s$, namely, $s\ge-\frac{1}{4}||x_0-x_1||^2$.

The point is that
\begin{equation*}
\langle x-x_0,x-x_1\rangle = \left\lVert x - \frac{x_0+x_1}{2}\right\rVert^2-\frac{1}{4}||x_0-x_1||^2\tag{1}
\end{equation*}
so we need $s+\frac{1}{4}||x_0-x_1||^2\ge 0$.

You can play with this at [Desmos:Circles1](https://www.desmos.com/calculator/46lufaqyu7)

To verify (1) this you simply compute:
\begin{align*}
\langle x-x_0,x-x_1\rangle&=||x||^2-\langle x_0 + x_1,x\rangle + \langle{x_0,x_1}\rangle
\end{align*}
and
\begin{align*}
\left\langle x - \frac{x_0+x_2}{2},x - \frac{x_0+x_2}{2}\right\rangle&=\langle x,x\rangle -\langle x_0+x_1,x\rangle + \frac{1}{4}||x_0+x_1||^2
\end{align*}
It is simple to show using similar computations, and it is known as the Parallelogram Law, that
$$\langle x_0,x_1\rangle = \frac{1}{4}(||x_0+x_1||^2-||x_0-x_1||^2)$$

Now back to Jonah's question we are looking at $p_i(x)=\langle x-x_i,x-x_i\rangle - r_i^2$ and $S_{r_i}(x_i)=Z(p_i)$ (the zero set). We want to add the two polynomials and understand the result geometrically. So First notice
\begin{align*}
p_0(x)+p_1(x)&=\langle x-x_0,x-x_0\rangle+\langle x-x_1,x-x_1\rangle-r_0^2-r_1^2\\
&=2\langle x - x_0,x - x_1\rangle + ||x_0-x_1||^2-r_0^2-r_1^2
\end{align*}
To see this just compute:
\begin{align*}
\langle x-x_0,x-x_0\rangle&+\langle x-x_1,x-x_1\rangle\\
&=||x||^2-2\langle x_0,x\rangle+||x_0||^2+||x||^2-2\langle x_1,x\rangle +||x_1||^2\\
&=2||x||^2-2\langle x_0+x_1,x\rangle+||x_0||^2+||x_1||^2\\
&=2||x||^2-2\langle x_0+x_1,x\rangle + 2\langle x_0,x_1\rangle+||x_0||^2-2\langle x_0,x_1\rangle + ||x_1||^2\\
&=2\langle x-x_0,x-x_1\rangle + \langle x_0-x_1,x_0-x_1\rangle\\
&=2\langle x-x_0,x-x_1\rangle + ||x_0-x_1||^2
\end{align*}
So the sphere produced by the algebraic sum is given by:
\begin{equation}
\langle x-x_0,x-x_1\rangle=\frac{1}{2}(r_0^2+r_1^2-||x_0-x_1||^2)\tag{2}
\end{equation}
You can play with this at [Desmos:Jonah Circles](https://www.desmos.com/calculator/sp7b0332q6)

As discussed above, this is the sphere
\begin{align*}
\left\lVert x- \frac{x_0+x_1}{2}\right\rVert^2&=\frac{1}{4}||x_0-x_1||^2+\frac{1}{2}\left(r_0^2+r_1^2-||x_0-x_1||^2\right)\\
&=\frac{r_0^2+r_1^2}{2}-\frac{1}{4}||x_0-x_1||^2
\end{align*}

Actually this last equation describes geometrically what the algebraic sum gives. The sum of $S_{r_0}(x_0)$ and $S_{r_1}(x_1)$ is the sphere $S_r(\bar x)$ where $\bar x=\frac{x_0+x_1}{2}$ and $r^2=\frac{r_0^2+r_1^2}{2}-\frac{1}{4}||x_0-x_1||^2$. So the center of the sum is simply the average of the centers. Let $d=\frac{1}{2}||x_0-x_1||$ be half the distance between centers, then
$$
r=\frac{r_0^2+r_1^2}{2}-d^2=\frac{1}{2}\left((r_0^2-d^2)+(r_1^2-d^2)\right)
$$

**I am not happy with this!**

**Geometric Interpretation** The sum of these two concentric spheres $C_1$ and $C_2$ is precisely the sum of the initial two circles.

You can play with this at [Desmos:Jonah Circles 2](https://www.desmos.com/calculator/ajsv3qhjdm)

The point is that the sum of two concentric spheres with radii $r_1$ and $r_2$ respectively, is simply described as the circle with the same center as the initial two and whose radius $r$ satisfies $r^2=\frac{r_1^2+r_2^2}{2}$.

**Geometric Interpretation (when spheres intersect)** In this case the sum of the circles has center being the midpoint on the line segment connecting the centers and whose radius is determined by any point in the intersection, since all points in the intersection must lie in the sum.