Complex linear transformation, show maps center of circle to center of circle 
If $L$ is a complex linear transformation with $L(z)=Az+B$, $A \neq 0$, $B \in \mathbb{C}$ and $L$ maps the circle $C_1$ onto the circle $C_2$ where
  $$C_1=\{z \in \mathbb{C}:|z-z_0|=R_1>0\} \text{ and } C_2=\{w \in \mathbb{C}:|w-w_0|=R_2>0\}$$
  then $L(z_0)=w_0$.

This is part of a problem I had in a complex analysis course a while back. I thought about maybe I could use properties of isometries but it looks like $L$ is only an isometry when $|A|=1$. I've tried to assume that $|L(z_0)-w_0|=\rho>0$ and I was thinking I could try to do this with cases, the first case being $0<\rho<R_2$ and the second case being $\rho \geq R_2$. 
For the first case, I was trying to come up with some kind of contradiction involving $w_0,f(z_0)$ and some point on the circle $C_2$, by using the triangle inequality. It looks like an arbitrary point on $C_2$ won't work so I was considering using some kind of point $v$ on the circle where if $w_0=u_0+iv_0$ then I would choose one appropriate $v=u_0+iv_1$, something where at least $w_0$ and the point on the circle share a common real or imaginary part. At this point I am just wondering if there is a more direct way to do this or other easier ways? 
 A: For a circle, $z=z_0+R_1 e^{i \phi}$.  Then
$$L(z) = A z+B = (A z_0 +B) + A R_1 e^{i \phi}$$
Accordingly, $R_2 = |A| R_1$ and $w_0 = A z_0 + B = L(z_0)$.
A: For posterity and in the hopes of further clarification or improvements I can make to understanding some of these things and to make my reasoning stronger I submit my solution.
As Ron Gordon points out, each $z \in C_1$ can be written as
$$z=z_0+R_1e^{it} \text{ for some $t \in [0,2\pi)$}.$$
Hence if we let $f=L|_{C_1}$ then for each $z \in C_1$,
$$f(z)=A(z_0+R_1e^{it})+B=L(z_0)+AR_1e^{it}.$$
By assumption $L(C_1)=C_2$ so
$$f(z)=L(z_0)+AR_1e^{it}=w_0+R_2e^{i\alpha(z)} \text{ where pick the $\alpha(z) \in [0,2\pi)$}.$$
Here is where my reasoning feels a bit hairy:
At this point I feel like I am throwing a few things together and saying more or less that as
$$L(z_0)+AR_1e^{it} \text{ and } w_0+R_2e^{i\alpha(z)}$$
are both like the same map, both mapping onto $C_2$ like a parameterization of sorts, that it follows $L(z_0)=w_0$ and $|AR_1|=|R_2|$ and so $|A|R_1=R_2$.
Please give me feedback if you have any to give, I am not sure how I can better state this so that it seems like what I am saying follows from what I have, in a better seeming way, stated.
