Is this proof that a Mobius transformation takes this specific form correct? Let $m ∈ Homeo^c(\overline {\mathbb C})$ and let $A$ be any non-trivial Euclidean circle in $\mathbb C$ with Euclidean centre $r$. Suppose that $m$ takes $A$ to a Euclidean circle $B$ so that $m(r)$ is the Euclidean centre of $B$.
Prove that there exist $a, b ∈ \mathbb C$ such that either $m(z) = az + b$ for all $z ∈ \mathbb C$ or $m(z) = a \overline z + b$ for all $z ∈ \mathbb C$.
Now from previous theorems I have that $Homeo^c(\mathbb C)=M \ddot ob$ and that every element of $M \ddot ob$ either has the form: 
$m(z)= \frac{az+b}{cz+d}$ $\quad$
 or $\quad$
$m(z)= \frac {a\overline z + b}{c \overline z + d}$
So what I believe I am really trying to prove is that if $m(r)$ is the Euclidean centre of $B$ then $c=0$ in which case $d$ will be absorbed into the coefficients $a$ and $b$.
My proof which I am unsure about is:
Let $\frac dc$ lie on $A$. Then $m(\frac dc)= \infty$ but as $B$ is a Euclidean circle this cannot be true so $c=0$.
Is this valid? I'm not sure as the question says $A$ is any non-trivial Euclidean circle so I am not sure if I can choose a specific circle and use proof by contradiction or if I have to do a proof for A unconstrained. I am also worried that I don't really use $r$ or $m(r)$ in my proof.
Alternatively does this work: 
Let $p$ and $q$ be 2 different points on $A$. Then $\frac{r-p}{r-q}=1$ as r is the centre of $A$.
If $m$ takes $r$ to the Euclidean centre of $B$ then $\frac{m(r)-m(p)}{m(r)-m(q)}=1$. 
Now sticking in the equation for $m$ in the above formula I believe after simplification I will get that $\frac{m(r)-m(p)}{m(r)-m(q)}= \frac{cq+d}{cp+d} \frac {r-p}{r-q} = \frac{cq+d}{cp+d}$ which is $1$ iff $c=0$. My problem with this second attempt is that I'm not sure whether $\frac {\overline r-\overline p}{\overline r-\overline q}=1$ given that $\frac {r-p}{r-q} = 1$ which is needed for the $m(z)= \frac {a\overline z + b}{c \overline z + d}$ case.
 A: I think both your proofs are wrong.
In your first proof you assume that $ \frac {d}{c} $ is on $A$ such assumptions are not to be made you need to proof it before you may use it.
The error is your second proof is more subtile you treat $r-p$ and $r-q$ as  distances while in fact they are complex numbers and more akin to a vector.
so while  $ \frac {|r-p|}{|r-q|} = 1 $  could be used. 
$ \frac {r-p}{r-q} = 1 $ only implies $ p = q$
An easier  solution is to make sure that $p,r,q $ are colliniar by equating $q$ with $2r-p$ 
Then for $B$ to be euclidean $m(r) -m(p) = m(q) -m(r)$ 
Note here that this is geometricly working with vectors not with distances, if $m(r) -m(p) = m(q) -m(r)$ then $p , q , r$ are collinear and $d(pr) = d(rq)$ 
Then using the identity 
$$ m(i) -m(j) = \frac{(ad-bc)(i-j)}{(ci+d)(cj+d)}$$ 
(proof left to you, remarkable nice short formula)
and $ q = 2r -p$
you get
$$\frac{(ad-bc)(r-p)}{(cr+d)(cp+d)} = \frac{(ad-bc)((2r-p)-r)}{(cq+d)(cr+d)} $$
$$\frac{(ad-bc)(r-p)}{(cr+d)(cp+d)} = \frac{(ad-bc)(r-p)}{(cq+d)(cr+d)} $$
$ ad-bc \not=0 $ otherwise $m$ is not a mobius transformation , $ r-p\not= 0 $ otherwise $ A$ is not a circle
$$\frac{1}{(cr+d)(cp+d)} = \frac{1}{(cq+d)(cr+d)} $$
$$\frac{cq+d}{(cr+d)(cp+d)(cq+d)} = \frac{cp+d}{(cq+d)(cr+d)(cp+d)} $$
$$\frac{c(q-p) }{(cr+d)(cq+d)(cp+d)} = 0 $$
$ q-p\not= 0 $  therefore $ c = 0 $
QED 
for the overline case there are Mobius transformations $n$ with $n(p) =q ,n(q) =p $ and $n(r) =r$
hopes this helps
