Show that $O$ traces out a circle in the pencil defined by $C$ and $M$

Let extend the original picture to get the following Here, $G$, $E$, and $F$ are midpoints of $BC$, $CD$, and $DA$ respectively.

Then $GE$ and $OF$ are both parallel to and equal to $\frac12BD$, $EF$ and $OG$ are both parallel to and equal to $\frac12AC$. Since $AC\perp BD$, $OGEF$ is a rectangle.

Since $OGEF$ is a rectangle, $OE$ and $GF$ are equal and intersect at their common midpoint $N$.

Next, since $\angle CME=\angle MCE=\angle MBH$, we have $$\begin{aligned}\angle HMB+\angle BMC+\angle CME &= \angle HMB+90^\circ + \angle MCE\\ &=\angle HMB+90^\circ + \angle MBH\\ &=180^\circ. \end{aligned}$$ Thus, $M$, $E$, and $H$ are colinear. In particular, $ME\parallel IO$, where $I$ is the center of the circle $c$.

On the other hand, the two small arc $CD$ and $AB$ add up to $180^\circ$, so $$CD^2+AB^2 = 4R^2,$$ where $R$ is the radius of $c$. Since $ME =\frac12 CD$, we have $CD = IO$. Thus $OMEI$ is a parallelogram, and so $N$ is the midpoint of $MI$.

Now, look at $\triangle MOI$, we have $$MO^2+OI^2=R^2.$$ So $NO$ is constant by the median length formula.

Lastly, we already showed that $E$, $M$, and $H$ are colinear, so $EH\perp HO$. Thus $H$ belongs to the circle with center $N$ and radius $NO$.