I have $n \in \mathbb{N}$ values lying in a real, circular domain of period $T \in \mathbb{R}^{+*}$:
$(\xi_i \in [0, T[^c)_{i \in \{1,\dots, n\}}$
I refer to the domain $C_T = [0, T[^c$ as a "circular domain" or a "torus" because the neighbourhood of $T$ is also the neighbourhood of $0$, so that if you go straight from $0$ to $.9\,T$ and keep going, you may end up back to $0$ etc.
How do I compute the mean of $(\xi_i)$ taking this circular structure into account?
For $n=2,\ \xi_1 = .21\,T,\ \xi_2=.81\,T$, a naive attempt yields:
$\displaystyle\frac{1}{n}\displaystyle\sum_{i=1}^{n}\xi_i = \frac{\xi_1 + \xi_2}{2} = \frac{.21\,T + .81\,T}{2} = .51\,T$
whereas $.01\,T$ is expected as a result, which is the actual medium point between $\xi_1$ and $\xi_2$ or their "center of mass".
[EDIT]: This Wikipedia page has finally been helpful :)
Note that $\xi_1$ is actually a class of equivalent points $\xi_1 = \{0.21 T + k T : k \in \Bbb{Z}\}$, itself and all of its cousins that are reached by traversing the circle $k$ times. Similarly, $\xi_2 = \{0.81 T + k T : k \in \Bbb{Z}\}$.
If instead of choosing the members of the class corresponding to $k=0$ (which is, in some way, arbitrary since we could be given any member of the class as a starting point) we choose the members with smallest absolute value, we get $\chi_1 = 0.21T$ and $\chi_2 = -0.19T$ and the usual mean of these is $\frac{\chi_1 + \chi_2}{2} = 0.01 T$.
The wiki page you mention in an answer (which could have been edited into your question) indicates that generally the circular mean is formed from the choice of members from each class so that the resulting usual mean is minimized. This is the content of "$\bar{\alpha} = \underset{\beta}{\arg\min} \sum_{j=1}^n d(\alpha_j,\beta)$, where $d(\varphi,\beta) = 1-\cos(\varphi - \beta)$" there.