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

Question

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.

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Note: I add an illustration here:

Answer 1

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 $NO$.

Answer 2

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.

Answer 3

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 $O$.