Denote by $G$ the centroid, by $M$ the midpoint of $AB$ and by $X$ the second intersection of $CM$ and the circumcircle $\omega$ of $ABC$. By homothety $\mathcal{H}(A,2)$ we learn that $GC = GX$. Then \[\frac{1}{4}AB^2 = p(M,\omega) = \frac13 CM \cdot \frac23 CM,\] thus the median $CM$ has fixed length, which describes the desired locus. Conversely, we see that it is also sufficient.

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