Unfamiliarity with complex geometry With some drawings, the answer is likely a line perpendicular to $AO$.

In the plane are given a circle with center $O$ and radius $r$ and a point $A$ outside the circle. For any point $M$ on the circle, let $N$ be the diametrically opposite point. Find the locus of the circumcenter of triangle $AMN$ when $M$ describes the circle.

If we let $O'$ be the circumcenter of $\triangle AMN$ and the projection of $O'$ unto $AO$ to be $P$, then the plan is to prove that $P$ is not dependent upon $\angle AON$. There is probably a cozy synthetic method, but I'd like to get some practice with complex numbers.
Let $\angle AON=\alpha,|AO|=L.$ Let $m=0,n=2,a=Le^{i\alpha}+1,$ such that the diameter of the circle is $2$. I know of a formula $\frac{xy(\overline{x}-\overline{y})}{\overline{x}y-x\overline{y}}$ to calculate the circumcenter of a triangle $XYZ$ when $Z=0.$ I'm not sure if my usage here is correct:
Substituting our values with $x=n=2$ and $y=a=Le^{i\alpha}+1$, we get $$O'=\frac{2(Le^{i\alpha}+1)(2-(Le^{-i\alpha}+1))}{2(Le^{i\alpha}+1)-2(Le^{-i\alpha}+1)}=\frac{1-L^2e^{2i\alpha}}{2L\sin\alpha},$$
where we get that second step from a denominator of $Le^{i\alpha}-Le^{-i\alpha}.$ That simplifies to $$O'=\frac{1-L^2(\cos2\alpha + i\sin2\alpha)}{2L\sin\alpha}.$$
To continue, we find $p$ as the foot of the altitude from $O'$ to $AO$. However, it turned out for me that I didn't get to anything meaningful because $p$ was dependent upon $\alpha$. I would write it here, but I'm worried that my result above is already incorrect. Is there anything I can fix, or does that look fine and that using the projection formula and some bash would get me to the result of $p$ being independent from $\angle AON$, and hence that $O'$ would be along the line $AO$?
 A: It is hard to understand what happens in the computations. (Starting from $m=0$ and $n=2$. If these are the afixes of $M,N$, then i expect them to move around a center.)
Here are some words and a complex numbers approach to the story.
For a point $Q$ in the plane i will denote by $z_Q\in\Bbb C$ the afix of $Q$.
Here is a picture for the convenience of the reader.


We will place $A$ in the origin, since we will use a formula which is nice to use when $A$ is in the origin.
The center $C$ of the circle, where the opposite points $M,N$ move on is taken on the real axis to simplify calculations, so let $c\in\Bbb R$ be the afix of $C$.
(I am not using the letter $O$ from the OP to avoid confusions.)
Let $O'$ be the circumcenter of $AMN$. Then we have:
$$
\begin{aligned}
z_A &= 0\ ,\\
z_C &= c\in\Bbb R\ ,\\
z_M &= c+re^{it}\text{ for some suitable }t\in\Bbb R\ ,\\
z_N &= c-re^{it}\ ,\\
z_{O'} &=\frac{z_Mz_N(\bar z_M-\bar z_N)}{\bar z_Mz_N-\bar z_N z_M}\\
&=
\frac 
{(c^2-r^2e^{2it})\Big(\ (c+re^{-it})-(c-re^{-it})\ \Big)}
{(c+re^{-it})(c-re^{it})-(c-re^{-it})(c+re^{it})}
\\
&=
\frac 1
{-2cr(e^{it}-e^{-it})}
(c^2-r^2e^{2it})\cdot 2re^{-it}
\\
&=\frac 1{c\cdot 2i\sin t}\color{blue}{(r^2e^{it}-c^2e^{-it})}\ .
\end{aligned}
$$
The quantity in the blue bracket is explicitly
$$ (r^2-c^2)\cos t + (r^2+c^2)i\sin t\ .$$
This shows that the real part of $z_{O'}$ is constant,
$$
\Re z_{O'}=\frac{(r^2+c^2)i\sin t}{c\cdot 2i\sin t}
=
\frac 1{2c}(r^2+c^2)\ .
\ .
$$
This corresponds to the "plan" from the OP, the projection of $O'$ on $AC$ (the real axis) is a fixed point, the point $P$ with $z_P=\frac 1{2c}(r^2+c^2)\in\Bbb C$.
$\square$
(The imaginary part involves the cotangent function, so that the whole perpendicular in $P$ on $AC$ is taken, not only a part of it.)
