In $\mathbb{R}^2$ every three points that are not colinear lie on a unique circle. Does this generalize to higher dimensions in the following way:

If $n+1$ element subset $S$ of $\mathbb{R}^n$ does not lie on any linear manifold (flat) of dimension less than $n$, then there is a unique $(n-1)$-sphere containing $S$.

If not, then what would be the proper generalization?

  • $\begingroup$ This result is true when $n=3$ and it is sufficient that the points are non coplanar. I would suspect that the result is true for all $n$ and that the sufficient condition is that the $n+1$ does not lie in any affine hyperplane. $\endgroup$ – C. Falcon Jun 26 '16 at 13:24

Hagen von Eitzen's answer gives a neat theoretical approach of this problem. However, I would like to expose a constructive and computational way to find the radius and center of the $(n-1)$-sphere determined by $n+1$ suitable points in $\mathbb{R}^n$.

Let $n$ be an integer greater than $1$ and let say $x_i:=(x_{i,j})_{j\in\{1,\cdots,n\}},i\in\{0,\cdots,n\}$ are $n+1$ given points. Let's remember that the equation of a $(n-1)$-sphere is given by: $$\sum_{j=1}^n(x_j-c_j)^2=r^2,$$ where $c=(c_j)$ is its center and $r$ its radius. Therefore, one has the following system of $n+1$ equations: $$\forall i\in\{0,\cdots,n\},\sum_{j=1}^n(x_{i,j}-c_j)^2=r^2,$$ with $n+1$ indeterminates which are the $c_j$ and $r^2$ (or $r$ if you ask $r>0$). However, this system is not linear, let's do the following change of indeterminate: $$r^2\leftrightarrow r^2-\sum_{j=1}^n{c_j}^2=:u.$$ Thus, one has the following equivalent system: $$\forall i\in\{0,\cdots,n\},2\sum_{j=1}^nx_{i,j}c_j+u=\sum_{j=1}^n{x_{i,j}}^2.$$ Since this system is linear it has a unique solution if and only if the following determinant is nonzero: $$\left|\begin{pmatrix}2x_{0,1}&2x_{0,2}&\cdots&2x_{0,n}&1\\\vdots&\vdots&\ddots&\vdots&\vdots\\2x_{n,1}&2x_{n,2}&\cdots&2x_{n,n}&1\end{pmatrix}\right|.$$ Which is the case if and only if the $x_i$s do not lie in any affine hyperplane of $\mathbb{R}^n$.

  • $\begingroup$ All the answers are great and exhibit an awesome diversity of perspectives. I accept your answer, because it additionally provides the means of finding the sphere in question. $\endgroup$ – Tom Jun 27 '16 at 16:37

Yes, if the $n+1$ points are in general position, which simply means that the $n+1$ points must not lie in a hyperplane.

We can proceed by induction: If $x_0,\ldots, x_{n}$ are our $n+1$ points in general position, then any $n$ of them, for example $x_0,\ldots, x_{n-1}$, certainly lie in a common $(n-1)$-dimensional hyperplane $H$. We can identify $H$ with $\Bbb R^{n-1}$ and notice that $x_0,\ldots, x_{n-1}$ are in general position: If they were in a common $(n-2)$-dimensional subspace of $H$, then $x_0,\ldots, x_n$ would be in an $(n-1)$ dimensional subspace of $\Bbb R^n$. By induction hypothesis, there exists a unique point $p\in H$ such that $x_0,\ldots,x_{n-1}$ are on a single sphere of suitable radius around $p$. Let $\ell$ denote the line in $\Bbb R^n$ that is normal to $H$ and passes through $p$. Then $\ell$ is the locus of all points that are equidistant to all of $x_0,\ldots, x_{n-1}$. Let $\ell'$ be the line through $x_n$ and $x_0$. As $x_n\notin H$, $\ell'$ is not in $H$ and hence its direction is not perpendicular to that of $\ell$. Let $H'$ be the hyperplane that bisects $x_0x_n$. Then $H'$ is perpendicular to $\ell'$ and so is not parallel to $\ell$. We conclude that $\ell$ intersects $H'$ in one and only one point $p'$. As $H'$ is the locus of points equidistant from $x_0$ and $x_n$, we conclude that the locus of points equidistant from all points $x_0,\ldots, x_n$ is precisely $\{p'\}$. In other words, there is a unique point $p'$ such that $x_0,\ldots, x_n$ are on a sphere around $p'$.

  • $\begingroup$ In case of an $n+1$ element subset $S$ of $\mathbb{R}^n$, isn't $S$ being in general position equivalent to it not lying on any hyperplane? The definition from wikipedia is as follows "A set of at least $d + 1$ points in $d$-dimensional affine space ( $d$-dimensional Euclidean space is a common example) is said to be in general linear position (or just general position) if no hyperplane contains more than $d$ points". $\endgroup$ – Tom Jun 26 '16 at 13:39
  • $\begingroup$ @Tom Hm, upon rereading, my wording seems to be suboptimal. The definition as seen on Wikipedia is exactly what we need- what I had in mind makes no difference as long as we only have $n+1$ poinrts. $\endgroup$ – Hagen von Eitzen Jun 26 '16 at 14:56

Why not just apply a circular inversion? If we have $p_0,p_1,\ldots,p_n\in\mathbb{R}^n$ in general position, we may consider $q_1,q_2,\ldots,q_n$ as the images of $p_1,p_2,\ldots,p_n$ under a circular inversion with respect to a unit hypersphere centered at $p_0$. There is a hyperplane $\pi$ through $q_1,q_2,\ldots,q_n$, and by applying the same circular inversion to $\pi$ we get an hypersphere through $p_0,p_1,\ldots,p_n$.
The uniqueness part is easy.


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