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?