Let $O$ be the center of the circle; WLOG assume that $OA\ge OB$. Let the circle intersect $AC$, $BC$ at $Q$, $P$, respectively. We need to prove that the length of the segment $PQ$ does not depend upon the choice of the circle; we will prove that $PQ=1$. Consider the translation above a vector $\overrightarrow{OC}$; let $P'$, $Q'$ be the images of $P$, $Q$, respectively. From the law of sines we get $$\frac{QQ'}{\sin \angle QCQ'}=\frac{Q'C}{\sin \angle Q'QC}=\frac{1}{\sin \angle 120^\circ}=\frac{1}{\sin \angle 60^\circ}=\frac{P'C}{\sin \angle P'PC}=\frac{PP'}{\sin \angle PCP'}.$$Since $PP'=OC=QQ'$ and both angles $QCQ'$, $PCP'$ are obviously acute, we obtain $\angle QCQ'=\angle PCP'$. Thus $\angle Q'CP'=\angle QCP=60^\circ$ which means that triangle $Q'P'C$ is equilateral, i.e. $P'Q'=1$. Since $PQ=P'Q'$, we are done. EDIT: Easier solution (without using translation): From the law of sines $$\frac{OC}{\sin \angle OPC}=\frac{OP}{\sin\angle OCP}=\frac{1}{\sin 60^\circ}=\frac{1}{\sin 120^\circ}=\frac{OQ}{\sin \angle OCQ}=\frac{OC}{\sin \angle OQC}.$$As a result we get $\angle OPC=\angle OQC$ (they are both acute), therefore $P,Q,C,O$ are concyclic. Hence $\angle POQ=\angle PCQ=60^\circ$, thus triangle $POQ$ is equilateral and $PQ=1$.