The locus of midpoints of chords which pass through point $P$

Question

Given a fixed circle, find the locus of midpoints of its chords
which include a fixed point $P$

Source: Hadamard's Geometry

My proof is below, to which I request verification and critique; in particular, which exposition is better?.

Solution: Let the fixed point be $P$, the circle's center be $O$, and the midpoint of $OP$ be $M$. Then the desired locus is the circle $\bigcirc M$ with center $M$ and radius $MP$.

Proof: Consider any chord $AB$ passing through $P$, and let $X$ be the midpoint of $AB$. Observe that $MX \cong MP$. For the ray from center $O$ through $X$ makes $\angle OXP$ a right angle and $\triangle OXP$ a right triangle. Then $XM$ is its median and $XM = \frac 1 2 OP = MP$. Thus $X$ is on $\bigcirc M$.

By a similar argument, the converse is true: Consider any point $X$ on $\bigcirc M$. By Thales Theorem, $\angle OXP$ is a right angle, and therefore $OX$ is the perpendicular bisector of the chord $AB$ containing $X$ and $P$, completing the proof.

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Below is an alternative exposition. Which is better proof writing?

Alternative Exposition

By Thales' Theorem and its converse, the locus of points $X$ such that $\angle OXP$ is a right angle is precisely the circle with center diameter $OP$.

A point $X$ is the midpoint of a chord $AB$ through $P$ iff $OX$ is the perpendicular bisector of $AB$. Thus any such midpoint $X$ will make $\angle OXP$ a right angle. Conversely, if $\angle OXP$ is a right angle, then $X$ is the midpoint of chord $AB$ through $P$, completing the proof.

Answer 1

Your basic method is sound:

1. Deriving $X$ from $A$ and $B$, establish that $OX \perp XA$ (equivalently, $XB$ or $XP$).
2. Recall that in a right triangle, the midpoint of its hypotenuse is equidistant from its vertices (i.e. it is its circumcentre).
3. Conclude that $X$ necessarily lies on $\bigcirc M$.
4. Conversely, deriving $X$ from $\bigcirc M$, extend $XP$ to a chord $AB$.
5. Show that $X$ is the midpoint of this chord, so lying on $\bigcirc M$ is also sufficient.

Looking at step 4, we can say that proof 1 does this directly. Proof 2 starts at $OX \perp XP$ and shows that $X$ lies on $\bigcirc M$ as a consequence. I think this is a reason to favour proof 1, although subjectively I actually preferred proof 2!

Possible improvements are twofold: you could account for edge cases, and you could clarify your reasoning.

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In the first line, allowing $O$ and $P$ to coincide is probably not worth worrying about, simply because it is trivial ($AB$ is a diameter and its midpoint is exactly $O$).

I think it is worth taking the time to consider the case where $X$ coincides with either $O$ or $P$, though, just to prove that all points on $\bigcirc M$ are valid. This means considering that $AB$ can be either a diameter of or tangent to $\bigcirc M$, so Thales’ theorem doesn’t apply.

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In the second line, I stumbled at this part of your first proof:

Observe that $MX \cong MP$. For the ray from center $O$ through $X$ makes $\angle OXP$ a right angle and $\triangle OXP$ a right triangle. Then $XM$ is its median and $XM = \frac 1 2 OP = MP$.

The second and third sentences comprise a proof of the first. But they also suffice as a proof of your overall statement. So the lemma “$MX \cong MP$” is not needed.

Your second proof starts with a statement about right angles, before establishing that such an angle exists. I think it works better in this order:

A point $X$ is the midpoint of a chord $AB$ through $P$ iff $OX$ is the perpendicular bisector of $AB$. Thus any such midpoint $X$ will make $\angle OXP$ a right angle. Conversely, if $\angle OXP$ is a right angle, then $X$ is the midpoint of chord $AB$ through $P$.

By Thales' Theorem and its converse, the locus of points $X$ such that $\angle OXP$ is a right angle is precisely the circle with diameter $OP$.

(Note also the deletion of “center” from “center diameter $OP$”.)

Like your first proof, here the second and third sentences (“Thus” and “Conversely”) are a restatement of the “iff” in the first sentence. I think it would be clearer if one or the other were replaced with a justification of this relationship between a chord and its perpendicular bisector. It needn’t be a complete proof, just a pointer from this specific case to the general rule. For example:

Since $O$ and $X$ are both equidistant from $A$ and $B$, $OX$ is the perpendicular bisector of $AB$. Thus $\angle OXP$ is a right angle, and conversely, if $\angle OYP$ is a right angle, then $Y$ is the chord’s midpoint.