Meaning of euclidean distance matrix's rank It is well known that in dimension $m$, an Euclidean distance matrix has rank less than or equal to m+2. Wiki also knows about it
Also another known fact is added there:

If the points $x_{1},x_{2},\ldots ,x_{n}$ are in general position, the
  rank is exactly $min(n, m + 2)$.

But wherever I read about it, the opposite statement is never mentioned, seemingly being wrong. So, let's say I've got a $n \times n$ Euclidean distance matrix and it really has got the rank equal to $m+2$. 
What would it mean? Are there any other definite opportunities for that matrix other than to contain points in general position? The reverse statement has not been proven or there are more options (possibly undiscovered yet)?
So, briefly speaking, what conclusions can be made from the fact that the Euclidean distance matrix has it's rank equal to $space \ dimension + 2$?
 A: Well, had to search on the topic for a while. Something concrete had been found in an old, but frequently cited alticle.
Theorem 6 there states:

All the null vectors of $D$ are also null vectors of $F$, and they
  satisfy $e^Tx_i = 0$. $F$ always has $s$ as a further null vector. If
  $rank(D) = r$ then the dimensionulity of $D$ is: 
$(i)$ $r-1$ iff $e^TD^-e\neq0$, in which case the generating points
  lie on the surface of a hypersphere, or 
$(ii)$ $r-2$ iff $e^TD^-e=0$, in which case the generating points
  cannot lie on the surface of a hypersphere and $D^-e$ is a further
  independent null vector of $F$.

...where $D$  is a EDM (Euclidean distance matrix) – real symmetric $n \times n$
matrix with elements $-\frac{1}{2}d_{ij}^2$ and with zero diagonal; the dimensionality of $D$ is the dimension of the space containing the
points that generate $D$; further details can be found in the original article.
(Gower, J. C., Properties of Euclidean and non-Euclidean distance matrices, Linear Algebra Appl. 67, 81-97 (1985). ZBL0569.15016.)
So, seemingly Wikipedia was not precise enough – general position in the case means "general" in terms of hypersphere. In other words, $rank(D) = r + 1$, if and only if the points generating $D$ lie on the relative boundary of an $r$-dimensional hypersphere.
