Let us fix $a=se^{i\psi}\in\Bbb C,\;r>0,\epsilon>0$.
I want to describe the set $$ A:=\left\{z\in\Bbb C\;:\;\left|\frac{\epsilon}z-a\right|<r\right\}\;\;. $$
First I notice that $$ \frac{\epsilon}z=a\;\;\Longleftrightarrow z=\frac{\epsilon}{s}e^{-i\psi}. $$ Call it $z_0$. Now, if we move in $A$, on the line of complex numbers of argument $-\psi$, the most distant elements from $z_0$ are $z_{\pm}:=\left(\frac\epsilon s\pm r\right)e^{-i\psi}$ (which are in $\partial A$ to be precise), so it follows that $$ \lambda e^{-i\psi}\in A,\;\;\lambda\in\left(\frac\epsilon s-r,\frac\epsilon s+r\right). $$ Moreover, for every such $\lambda$, there exists $g(\lambda)>0$ such that $$ \lambda e^{-i\phi}\in A\;\;\Longleftrightarrow \phi\in\left(\psi-g(\lambda),\psi+g(\lambda)\right)=:I_{\lambda}\;. $$ It is clear that:
-$g$ is continuous (and I'd say smooth)
-$g(\lambda)\to0^+$ as $\lambda\to\frac\epsilon s\pm r$
-it reaches its maximum at the midpoint $\lambda_0=\frac\epsilon s$
-it's symmetric with respect to $\lambda_0$
Hence $A$ can be described as follows:
$$ A=\left\{\lambda e^{-i\phi}\;:\;\lambda\in\left(\frac\epsilon s-r,\frac\epsilon s+r\right),\; \phi\in\left(\psi-g(\lambda),\psi+g(\lambda)\right)\right\}\;\;. $$
My suspect is that $g$ increases till $\lambda_0$ and then decreases to $0$ in a way that the surface spanned is exactly a circle, namely $$ A=\left\{\left|z-\frac\epsilon a\right|<r\right\}\;. $$ What can I do to get more information on $g$ to prove/disprove this last equality?
$W = \{w \in \mathbb C: |w - a| = r\}$ is the circle of radius $r$ centred at $a$. It is the image of $Z = \{z \in \overline{\mathbb C}: |\epsilon/z - a| = r\}$ under the inversion $f: z \to \epsilon/z$, where $\overline{\mathbb C}$ is the Riemann sphere (the extended complex plane). That being a Möbius transformation, it takes circles and lines into circles and lines. To make sure it is a circle and not a line, we just need to make sure it doesn't contain the point at $\infty$, i.e. that $|a| \ne r$. Now $A = \{z \in \mathbb C: |\epsilon/z - a| < r\}$ is either the inside or the outside of the circle $Z$. To check which it is, we just need to check what happens to $\infty$ under the transformation. What we need for it to be the inside of the circle is that it doesn't contain a deleted neighbourhood of $\infty$, i.e. that $|a| > r$.