Let $m \in \text{Homeo}^{C}(\bar{\mathbb{C}})$ (the group of homeomorphisms of $\bar{\mathbb{C}}$ taking circles in $\bar{\mathbb{C}}$ to circles in $\bar{\mathbb{C}}$, where $\bar{\mathbb{C}}=\mathbb{C} \cup \lbrace\infty\rbrace$). Let $A$ be any non-trivial Euclidean circle in $\mathbb{C}$ with Euclidean centre $c$. Suppose that $m$ takes $A$ to a Euclidean circle $B$ so that $m(c)$ is the Euclidean centre of $B$.
How would I go about proving that there exist $a,b \in \mathbb{C}$ such that $\forall z\in \mathbb{C}$ either $m(z)=az+b$ or $m(z)=a\bar{z}+b$?
I'd be very grateful for any guidance you can offer.
I think the problem is not perfectly posed, since any möbius transformation maps circles into circles in the Riemann sphere, which I suppose is what you mean by $\overline{\mathbb{C}}$. I'm not good in finding clever and short solutions to problems, but I suggest a way by brute force.
Take the unit circle $C$ centered at the origin and suppose that $m(0)=\tau$ and the radius of the image of $C$ by $m$ is $r$, so by translation and dilatation of $m$, we can assume that $m(C)=C$. Take now the circle $1+C$, i.e. translate the unit circle one unit to the right, what is $m(1+C)$? The center of $1+C$, that is 1, lies in $C$, hence $m(1)\in C$ and $0\in 1+C$, so $m(1+C)$ is a unit circle centered at $C$. By a rotation $e^{2\pi i\theta}$ we can assume that $m(1)=1$. Now consider the circle $2+C$ a try to convice yourself that $m(2)=2$, (use that $2+C$ contains the center of $1+C$, that $2\in 1+C$ and that $2+C$ is tangent to $C$) and in general $m(n)=n$, for $n\in\mathbb{Z}$.
If we consider the circle $\frac{1}{2}+\frac{1}{2}C$, i.e. a circle of ridius $\frac{1}{2}$ centered at $\frac{1}{2}$, then this circle contains the centers of $C$ and $1+C$ and is tangent to both of them, hence the only possibility is that $m(\frac{1}{2}+\frac{1}{2}C)=\frac{1}{2}+\frac{1}{2}C$ and then $m(\frac{1}{2})=\frac{1}{2}$. Following the same pattern, at the end you get that $m(t)=t$ for $t$ in a dense subset of $\mathbb{R}$, and by continuity $m$ is the identity in $\mathbb{R}$.
Since the circles $C$ and $1+C$ intersect at the points $e^{i\pi/3}$ and $e^{-i\pi/3}$, then $m(e^{i\pi/3})=e^{\pm i\pi/3}$, so by conjugation we can assume that $m(e^{i\pi/3})=e^{i\pi/3}$ and going on playing the same game with circles, we eventually find out that $m$ is the identity in a dense subset of $\mathbb{C}$.
I hope it helps.