This was nice and simple it was just letting $O_2$ be the circumenter of $ABF$ and prove $O_1$ is obtained by ortating $O_2$ and $O_2$ varies over a line and u need to prove rotation of a line is also a line.

Let $BE$ meet $AFC$ at $K$. Claim $: AK$ is tangent to $ABC$. Proof $:$ Note that $\angle KAC = \angle KFC = \angle FBC + \angle FCB = \angle B$. Note that since $\angle C$ is fixed then $K$ is fixed and so perpendicular bisector of $AK$ is also fixed. Note that circumcenter of $AFC$ lies on this fixed line so the locus of circumcenters is the perpendicular bisector of $AK$.