Q:

How many points define a sphere of unknown radius?

In $\mathbb{R}^n$, is it true that any given $n+1$ points that don't lie in a hyperplane lie on one and only one sphere ?
I think it's true in $\mathbb{R}^2$ and I was trying to work it out for $\mathbb{R}^3$.
An equivalent formulation of the question is :
For any given points $a_0, \dots , a_n\in \mathbb R ^n$ not contained in any hyperplane, does there exists a radius $r$ such that the set of spheres $\lbrace S(a_i, r)\rbrace_{0\leq i \leq n}$ has non empty intersection ?
Any point in that intersection is then the center of a/the sphere containing all of the points.
If the question depends on the chosen norm, then I suggest considering the euclidean norm.

A:

Yes they do, here's one proof (for the Euclidean norm):
You can assume, by translating, that the points are $\mathbf{0},\mathbf{x}_1,\ldots,\mathbf{x}_n \in \mathbb{R}^n$. Then you're trying to find a point $\mathbf{p}$ which is equidistant from all the other points, i.e. such that $||\mathbf{p}||^{2} = ||\mathbf{p} - \mathbf{x}_i||^{2}$ for all $i$. This simplifies to $2\mathbf{p} \cdot \mathbf{x}_i = ||x_i||^{2}$ This can be viewed as a linear system: $A\mathbf{p} = \mathbf{v}$, where $A$ is the matrix whose $i$th row is $2\mathbf{x}_i^{T}$, and $\mathbf{v}$ is the column vector whose $i$th entry is $||x_i||^{2}$. The fact that the $\mathbf{x}_i$ and $\mathbf{0}$ don't all lie in a hyperplane implies $\mathbf{x}_i$ are linearly independent, meaning $A$ is invertible and hence the system has a (unique!) solution.

