Given the bilinear transformation $$w=\frac{z+2i}{2iz-1}.$$ What is the preimage of $\{w \mid |w|=1\}$, in other words, what does $z$ (its inverse) map $\{w \mid |w|=1\}$ to?
Since this is a Mobius transformation i know that it must map a circle, but which I am not sure, and how to find the equation either. I obviously know that $$z=\frac{w+2i}{2iw-1}$$
$$w=\frac{z+2i}{2iz-1}, |w|=1$$
$$\implies \left|\frac{z+2i}{2iz-1}\right|^2=1 \implies |z+2i|^2=|2iz-1|^2$$
$$\overset{|z|^2=zz^*}{\implies} |z|^2+2iz^*-2iz+4=4|z|^2-2iz+2iz^*+1$$
$$\implies 3=3|z|^2 \implies |z|^2=1$$
So, the preimage of the unit circle under this map is, in fact, the unit circle, i.e. $w=w(z)$ preserves the unit circle.
Let $$f(z) = w=\frac{z+2i}{2iz-1}; \quad f^{-1}(w) = z=\frac{w+2i}{2iw-1}.$$
Here's one approach. You mentioned that you know Möbius transformations send the set of all circles and lines to itself. So, letting $U = \{w : |w| = 1\}$, you know that $f^{-1}(U)$ is either a line or a circle. You just need to figure out which of the two it is.
It is easy to tell the two apart: $3$ points on a line are collinear; $3$ points on a circle are not. Thus, we pick $3$ points $1, i, -i \in U$ and compute:
\begin{align*} f^{-1}(1) &= [\text{compute me}] \\[7pt] f^{-1}(i) &= \frac{i+2i}{-2 - 1} = -i \\[7pt] f^{-1}(-i) &= \frac{-i + 2i}{2 - 1} = i \end{align*}
Now it is clear that $f^{-1}(U)$ is a (line/circle), and moreover, it is the (line/circle) ...
In this problem, many stars aligned. In general, it won't be as nice to figure out what circle we're dealing with (lines are always nice). But it's a general fact that there's a unique circle through any $3$ non collinear points. It is a good exercise to find the center and radius for a circle going through $3$ non collinear points, or at least see why there is a unique such circle (draw a picture! Consider the perpendicular bisectors of line segments joining two pairs of points).