A Mobius transformation of the plane takes $z \mapsto \frac{az+b}{cz+d}$. These are known to take circles to circles, but given an explicit circle, how do we compute the radius.
Let's parameterize our circle by $z(t) = z_0 + r e^{2\pi i n t}$. What is the radius and center of the image circle?
$$ \frac{a\,z(t)+b}{c\,z(t)+d}$$
I am looking for a computational proof that the image is a circle so I can find the (Euclidean) radius and center.
Source: Wikipedia
In my application, I have an approximate circle $\{ z_0 + e^{2\pi i n t}: t \in \frac{1}{N}\mathbb{Z} \}$ where $N$ is a large number. If we act the Mobius transformation pointwise, these spaces will no longer be evenly spaced out. So I decided it's better to compute the Euclidean center and radius if possible.
I'm not sure this will fully satisfy you but I might as well write it down.
Instead of centering your original circle $\Gamma$ on the origin, you can center your Mobius transform so that it becomes $z\mapsto \frac{a}{z}+b$. This reduces the problem to finding the image of a given circle under complex inversion, which is much easier.
Your image circle $\Gamma'$ is the symmetric reflection of the image of $\Gamma$ under $\phi:z\mapsto \frac{1}{\bar{z}}$. It is easily checked that $\phi$ is the transformation of $\mathbb{R}^{2}$ that associates to a vector $x$ the vector $\frac{x}{||x||^{2}}$. By a symmetry argument (look at the tangents to $\Gamma$ passing throught the origin), the origin, the center of $\Gamma$ and the center of $\Gamma'$ are colinear. This allows you to calculate your radius and your center.
With all this you can do the computations, but I don't really feel like explicitly writing them down... You can also turn the above discussion into a proof that the image is indeed a circle: it is enough to prove that $\phi$ sends circles not passing through the origin into other circles, and this can be done with elementary geometry (anti-similar triangles) by showing that $\phi$ preserves angles between three points.