Given a triangle $T$, we can define a transfinite sequence $(\mathsf{CS}_\alpha(T))_{\alpha\in\mathsf{Ord}}$ of "circumsets" as follows:
$\mathsf{CS}_0(T)=T$,
$\mathsf{CS}_{\alpha+1}(T)=\mathsf{CS}_\alpha(T)$ $\cup $ the set of all circumcenters of nondegenerate triangles with vertices in $\mathsf{CS}_\alpha(T)$, and
$\mathsf{CS}_\lambda(T)=cl(\bigcup_{\alpha<\lambda}\mathsf{CS}_\alpha(T))$ for $\lambda$ limit (where $cl(A)$ is the closure of $A$ in the usual topology on $\mathbb{R}^2$).
Let the circumordinal of $T$, $\mathsf{co(T)}$, be the smallest ordinal $\alpha$ such that $\mathsf{CS}_\alpha(T)=\mathsf{CS}_{\alpha+1}(T)$. It's easy to show that $\mathsf{co}(T)<\omega_1$: given points $a,b,c\in \mathsf{CS}_{\omega_1}(T)$ we can find sequences $(a_i)_{i<\omega},(b_i)_{i<\omega},(c_i)_{i<\omega}$ of points in earlier $\mathsf{CS}$-levels converging to $a,b$, and $c$ respectively, but since $\omega_1$ is regular this means that $a,b,c\in\mathsf{CS}_\alpha(T)$ for some $\alpha<\omega_1$.
Are there arbitrarily large countable circumordinals? More ambitiously, which countable ordinals are circumordinals?
(The more ambitious question seems nontrivial even in the finite case.) I have only been able to find a small amount of existing work about iterated circumcenter processes (e.g. this MO discussion or this student(!) paper), none of which seems directly related, but I'm not an expert so I could easily be missing something obvious.
I was having a silly moment: it's easy to show that $\mathsf{co}(T)\le\omega$ always.
Specifically, let $X=\bigcup_{n<\omega}\mathsf{CS}_\alpha(T)$. Then $X$ is "a circle-center set" - the circumcenter of every triangle with vertices in $X$ is itself in $X$. But the map sending a triangle to its circumcenter is appropriately continuous, so in fact the closure of $X$ is also circumcenter-closed. (A bit more rigorously: given noncollinear points $a,b,c\in cl(X)$ we can find sequences $a_i\rightarrow a, b_i\rightarrow b, c_i\rightarrow c$ of points in $X$. Letting $d_i$ be the circumcenter of $\triangle a_ib_ic_i$, we have $d_i\in X$. But the sequence of $d_i$s converges to the circumcenter of $\triangle abc$.)
In the comments above, Greg Martin mentioned a paper showing that the only closed circle-center set is $\mathbb{R}^2$. Combined with the above this means that $\mathsf{co}(T)=\omega$ for all $T$.