Show that $O$ traces out a circle in the pencil defined by $C$ and $M$ 
Let $C$ be a circle and $M$ any point inside the circle. Consider a moving chord $AB$ of the circle which stuntedness a right angle at $M$. Let $O$ be the midpoint of this chord and let $H$ be the altitude from $M$ on $AB$ so that the angle of $BHM=90$degrees.
If we consider $M$ as a degenerate circle of radius zero, show that $O$ traces out a circle in the pencil defined by $C$ and $M$ as the chord $AB$ moves around the circle. Furthermore, show that $H$ lies on the same circle.
So I've already shown that $P_c(O)=-P_m(O)$. I can somewhat see that $O$ traces out a circle inside of $C$ with the point $M$ in the circle, but i don't really know how to explicitly show that $O$ traces out the circle using this power property.
The answer below are very insightful but not exactly what i'm looking for. 
I just learnt pencils of circles and i'm having a tough time really understanding it and how it applies. Especially how to show that it traces out a circle in the pencil. 

Note: I add an illustration here:

 A: 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 $NO$.
A: 
From the question, I think the co-ordinates of M should be some given quantities. WLOG, we can also assume that C’s equation is $x^2 + y^2 = R^2$ (for some known $R$) with center at $O(0, 0)$. 
$OM$, when extended, will be the line $L$ whose equation can be found. $L$ will cut $C$ at $X$ and $Y$. Their co-ordinates can be found by solving $L$ and $C$.
$L’$ is normal to $L$ and passes through $M$. The equation of $L’$ can be found. Solving $L’$ and $C$ will give us the co-ordinates of $B_1$ and $B_2$.
If $A$ is at $X,$ then B will be at $B_1$ or $B_2$ (and vice versa). Let $P$ be the midpoint of the chord $AB_1$. $Q$ is similarly defined. $R$ and $S$ will be similarly defined if $A$ is at $Y$. Note that the co-ordinates of $P, Q, R$ and $S$ can all be found.
In the added picture of the post, the required locus is circle. By midpoint theorem, it is not difficult to prove that $PQRS$ is a rectangle. Hence, they are con-cyclic points of a circle. That circle is exactly the required because $P, Q, R, S$ are particular four points of the circle.
Using the co-ordinates of P and R as endpoints of a diameter of the circle, we get the equation of that circle.
A: Call $I$ the center of the circle and pick two points $O$ and $M$ in the circle. We want to know when we can find $A,B$ on the circle such that $O$ is the middle of $[AB]$ and $AMB$ is a right angle. 
Looking at $O$,
there is only one chord $[AB]$ whose midpoint is $O$, and we have $r^2 = IO^2 + OA^2$.
Next, we need $M$ to be on the circle of diameter $[AB]$, so we want $MO = OA$.
Putting this together, we obtain the equation $r^2 = IO^2 + MO^2$, or $(IO^2-r^2) + (MO^2) = 0$.
But $IO^2-r^2 = 0$ is the equation of the circle, and $MO^2 = 0$ is the equation of the degenerate circle-point at $M$. Since we are looking at a linear combination of those two, the locus of $O$ is indeed a circle in the pencil containing $C$ and $M$.
I believe the other other answer shows that $H$ has the same locus as $O$.
