Let $O$ be the circumcenter, $N$ be the midpoint of $\overline{AH}$, $M$ be the midpoint of $\overline{BC}$. Note that $\overline{NM}$ is a diameter of the nine-point circle of $\triangle ABC$, so $NM = R$ which is fixed. Also $AH = 2OM$ is fixed as well. This pretty much implies the conclusion. The two fixed circles are the ones centered at $M$ with radii $NM - AN = R-OM$ and $NM+AN = R+OM$, which are the circles centered at $M$ tangent to $(ABC)$. see here for more discussion