Möbius transform which completely preserves circles (how to map a circle?) (remmert theory of complex function) 
I am trying to solve this exercise, however it seems impossible because I don't know how to map a circle, and I will be very thankful if somebody points out to me:  

Given a Circle $C$, is it possible to show that for $a_{1},b_{1}\in \mathbb{C}\backslash C $ there exists a Möbius transformation $$M(z)= \frac{az+b}{cz+d}$$
which fulfills: $$M(C)=C ; M(a_{1})=b_{1}$$

So we can write conditions: $M(a)=b \Rightarrow aa_{1}+b= b(ca_{1}+d)$
How does one map a circle? I thought that a circle in $\mathbb{C}$ needs 3 points to be uniquely determined , so we can put: $z_{1},z_{2},z_{3} \in C$
and the condition is the same as for the a to b mapping: $$az_{k}+b=z_{k}(cz_{k}+d), k=1,2,3 \in \mathbb{N}$$
It seems that there isn't much more one can do with this approach, so I think it is not the right one. What is the right approach?
 A: In your question you've taken the condition "$M(C)=C$ " and misunderstood it to mean that it leaves every single point on the circle unchanged, so that $M(z_{1,2,3})=z_{1,2,3}$ respectively. This interpretation is false. I provided a counterexample in my comment, but you misunderstood the point of the comment: the map $z\mapsto iz$ is a Möbius transformation that maps the unit circle to the unit circle (hence it satisfies the condition $M(C)=C$) however $M(z)=z$ is false for every single particular element $z$ on the unit circle. The point is that your interpretation of the condition does not make sense so you cannot build an approach off of it. I am not saying that pure rotations are the only transformations preserving circles; they are not.
The key to understanding this problem is geometry - more specifically, spherical geometry. We will first prove a particular case of the proposition, and then show this entails the full proposition.
Lemma. For any $u,v\in\mathbb{C}$, there is a Möbius transformation $\varphi$ such that $\varphi(u)=v$ and $\varphi(\mathbb{R})=\mathbb{R}$.
Visual Proof. The isometries of the real line are affine transformations $x\mapsto ax+b$, so we need to see that such a transformation is sufficient. Translations have an obvious interpretation of moving everything horizontally but not vertically, while real dilations will act on complex $z$ as dilations of $z$ on the ray extending from the origin to $z$. The line between the origin and $u$ will take on every possible $y$ (imaginary) value, so we can perform a real dilation on $u$ until it matches $v$ in height; on top of that we just need to translate the appropriate horizontal distance to get to $v$. With these two actions composed together we have our affine transformation, as desired.
Proposition. For any (generalized) circle $C$ and $u,v\in\mathbb{C}-C$, there is a Möbius transformation $\varphi$ such that $\varphi(u)=v$ and $\varphi(C)=C$.
Proof. Fix a Möbius transformation $\psi$ that maps $C\to\mathbb{R}$. By our lemma there is a transformation that preserves the real line but maps $\psi(u)$ to $\psi(v)$; call it $\rho$. Then our desired map is $\psi^{-1}\circ\rho\circ\psi $.
This write-up requires a bit of expansion in order to be as comprehensive as I'd like it to be, and I'll get to it next thing I do on MSE, but right now I have to go to sleep and catch some ZZZ's! (Also, this is basically of an elaboration on Blatter's answer and clarification of my comment.)
A: A hint: You are only asked to show that such an $M$ exists. Therefore you may assume that $C$ is some special circle to begin with. When $C$ is the real axis it is easy to write down a suitable $M$.
