First we'll prove that the chord $AB$ has a constant length. Consider $\angle BPA$. We know that $\angle QCP$ and $\angle PBQ$ are constant because they are subtended by constant chord $PQ$, but $\angle QCP + \angle PBQ = \angle BPA$, so $AB$ subtends a constant angle, hence its length stays constant as well. We conclude that $\frac{c}{\sin C} = 2R$ is constant, where $R$ is the circumradius. Therefore the circumradius of $ABC$ is constant, and it is now easy to realize that the locus of the circumcenters is an arc.

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