Consider distinct points $A,B$ and a line $\ell \perp AB$. For any point $C \in \ell$, form the circle $\circ ABC$, and take $C'$ as the point on $\circ ABC$ diametrically opposite $C$. What is the locus of such points $C'$?
Source: Hadamard's Geometry
My solution is below, to which I request verification, feedback, and suggestions. In particular: Is the exposition clear? How could it be improved?
Solution: Let line $p$ be the perpendicular bisector of $AB$. The desired locus is the reflection of $\ell$ over $p$, which we call line $\ell'$.
Proof: For any $C$ chosen, $AB$ is a chord of $\circ ABC$, and therefore the center $O$ of $\circ ABC$ is on $p$. Recall that the reflection $C'$ of any point $C$ over any point $O$ on line $p$ will have the same distance from line $p$ as $C$ itself has (as easily shown by AAS), and hence be on line $\ell'$. Conversely, reflecting any point $C'$ on $\ell'$ will yield an appropriate point $C$, completing the proof.
Update: The problem as written isn't clear if $C$ may be the unique point $\ell \cap AB$ (in which case we form not a circle but a line). If this is not allowed, we must also disallow the unique point $\ell' \cap AB$ as well.