Question concerning a possible constant sum of squares in a circle Assume a circle, of some radius $r$ say $r=1$. Consider a fixed $n$, say $n=30$, of points on the circumference of the circle. That $n$ points $(x_k,y_k)_{k=1,\ldots, n}$ define a mean in the interior of the circle, call this point m $(x_m,y_m) $.
Now we compute the "circular variance" , just the mean-of-squares of the distances
$$ \mathrm{msq}=\frac1n\sum_{k=1}^n ((x_k-x_m)^2+(y_k-y_m)^2) $$
Now we assume another set of $n$ points $(x'_k,y'_k)_{k=1,\ldots, n}$, which are selected such that they have the same mean. Is the circular variance the same  $\mathrm{msq}'=\mathrm{msq} $ ?
I've seen, that the two extreme situations where the $n$ points are accumulated at the green positions in the graph and where they are accumulated at the magenta positions in the graph( (where the fat red point indicates the mean)) , the msq-values are identical. I could try to program a routine to test the question approximately by brute force, but perhaps there is an analytic argument?
 
[update]: Michael Hardy's anwer solves this neatly. I find it a very nice observation, that that "circular variance" depends only on the mean of the points and is thus constant for all resulting means which lie on the same circle inside the original circle.
So with the mean laying on the inner circle with radius $\small r_m$ we get $$\small \operatorname{msq}=1-r_m^2$$
 Possibly worth an entry in the wikipedia? (in some "special points/geometric relations of a circle" - section , don't know the actual name for the existing collection) Btw, does this allow also a generalization to a continuous formulation instead of "n discrete points", say via integrals?
 A: Letting $\bar x$, $\bar y$ be the average $x$- and $y$-coordinates, this "variance" is
$$
\sum_{k=1}^n ((x_k-\bar x)^2+(y_k-\bar y)^2)= \left(\sum_k (x_k^2+y_k^2)\right) - 2n(\bar x^2 + \bar y^2) + n(\bar x^2 + \bar y^2)$$
$$
= n - n(\bar x^2 + \bar y^2),
$$
and this clearly depends on the $x$s and $y$s only through the averages.
Later edit: The usual 2-dimensional counterpart of the variance is the nonnegative-definite matrix
$$
\mathbb{E}\sum_{k=1}^n \begin{bmatrix} x_k^2 & x_k y_k \\ x_k y_k & y_k^2 \end{bmatrix}.
$$
The "variance" in this question is the trace of that matrix.  In many applications, it makes no sense to speak of this matrix as having a trace at all, because, for example, $x_i$ may be measure in dollars and $y_i$ in centimeters, and it is not at all unusual for that sort of thing to happen.  What if one changes the units in which $y_i$ is measured from centimeters to meters?  However, since in this problem we have the constraint $x_k^2+y_k^2=1$, that problem doesn't seem to arise.
Now notice that although the trace of this matrix depends on the $(x,y)$s only through their average, nonetheless the off-diagonal entries of the matrix depend on more than that; they depend on the correlation.
Still later edit: Suppose $F$ is a probability distribution supported on the circle of unit radius centered at $(0,0)$, and $X$ is a random variable with that distribution.  Then we have
$$\mathbb{E}(X)=\mu=\text{some point in the disk}.$$
Then
$$
\mathbb{E}(\|X-\mu\|^2) = \mathbb{E}(\|X\|^2) - 2\mu\cdot \mathbb{E}(X) + \|\mu\|^2 = 1-2\|\mu\|^2+\|\mu\|^2 = 1 - \|\mu\|^2.\tag{1}
$$
A theorem of elementary geometry that I remember learning in school when I was about 16 says that for two chords $AB$ and $CD$ of a circle intersecting at $P$, the products of lengths $AP\cdot PB$ and $CP\cdot PD$ are equal.  That is seen to be simply a special case of (1), when we remember that the variance of a probability distribution supported on a set of two points is just the product of the distances from those two points to their suitably weighted mean.
So we have a novel proof, and a generalization, of that theorem of elementary geometry.
