It is possible, but it requires some grunt work. Let $w_{z_0}(r, \theta) = \frac{1}{z_0+r e^{i\theta}}$.
We deal with a simple case first: If $z_0 = 0$, then $w_0(r,\theta) = \frac{1}{r}e^{-i\theta}$, which is a circle.
Now reduce the problem to a simpler one. Let $w_{z_0}(r, \theta) = \frac{1}{z_0+r e^{i\theta}}$, and suppose $z_0 = |z_0| e^{i \zeta}$. Then we have $w_{z_0}(r, \theta) = \frac{1}{z_0}\frac{1}{1+\frac{r}{z_0} e^{i\theta}} = \frac{1}{z_0}\frac{1}{1+\frac{r}{|z_0|} e^{i(\theta-\zeta)}} = \frac{1}{z_0} w_1(\frac{r}{|z_0|}, \theta-\zeta) $. So, we will analyze $w_1$ first, and then generalize to arbitrary $z_0 \neq 0$. We have $w_1(r, \theta) = \frac{1}{1+r e^{i\theta}} = \frac{1+r e^{-i\theta}}{1+2 r \cos \theta + r^2} = \frac{1+r \cos \theta - i r \sin \theta}{1+2 r \cos \theta + r^2}$.
Note that if $r=1$, then $w_1(1,-\pi) = \infty$, so we expect $\theta \mapsto w_1(1,\theta)$ to be a line. To see this, look at $w_1(1,\theta) = \frac{1}{2} \frac{1+ \cos \theta - i \sin \theta}{1+\cos \theta } = \frac{1}{2} (1 - i \frac{\sin \theta}{1+\cos \theta })$, so $\theta \mapsto w_1(1,\theta)$ is a line parallel to the imaginary axis.
Now suppose $r\neq 1$. Since we expect $\theta \mapsto w_1(r,\theta)$ to be a circle, we should look for the center first. If we have two points on a circle, we know that the perpendicular bisector of the segment joining the points passes through the center. Now note that $w_1(r,0) = \frac{1}{1-r}$, $w_1(r,\pi) = \frac{1}{1+r}$, hence the real part of the center must line on the line $\text{Re}\, z = \frac{1}{2} (\frac{1}{1-r} + \frac{1}{1+r}) = \frac{1}{1- r^2}$. Similarly, since $w_1(r,\frac{\pi}{2}) = \frac{1-i r}{1+r^2}$, $w_1(r,-\frac{\pi}{2}) = \frac{1+i r}{1+r^2}$, we see that the center must lie on the line $\text{Im}\, z = \frac{1}{2} (\frac{1-i r}{1+r^2} + \frac{1+i r}{1+r^2}) = 0$. So, we expect the center to be $\frac{1}{1- r^2}$. To confirm, look at $|w_1(r,\theta) - \frac{1}{1- r^2}|$:
\begin{eqnarray} |w_1(r,\theta) - \frac{1}{1- r^2}| &=& \frac{1}{|1-r^2|}\left| \frac{(1-r^2)(1+r \cos \theta - i r \sin \theta)}{1+2 r \cos \theta + r^2} - 1 \right| \\ &=& \left| \frac{(1-r^2)(1+r \cos \theta - i r \sin \theta)- (1+2 r \cos \theta + r^2)}{(1-r^2)(1+2 r \cos \theta + r^2)}\right| \\ &=& \frac{|r|}{|1-r^2|} \left| \frac{-(2r+(1+r^2)\cos \theta) - i (1-r^2) \sin \theta}{1+2 r \cos \theta + r^2}\right| \\ &=& \frac{|r|}{|1-r^2|} \frac{\sqrt{(2r+(1+r^2)\cos \theta)^2+((1-r^2) \sin \theta)^2}}{1+2 r \cos \theta + r^2} \\ &=& \frac{|r|}{|1-r^2|} \end{eqnarray} Hence $w_1(r,\theta)$ lies on a circle $\Gamma$ of radius $\frac{|r|}{|1-r^2|}$ centered at $\frac{1}{1- r^2}$.
To see that the map $\theta \mapsto w_1(r,\theta)$ is onto $\Gamma$, we note that the map $\phi: \mathbb{C}_\infty \to \mathbb{C}_\infty$ given by $\xi(z) = \frac{1}{z}$ is a homeomorphism (continuous with a continuous inverse). If we let $D=\{z||z|=1\}$, we note that $D^C$ is disconnected, and also $w_1(r,[0,2 \pi]) = \xi(D) \subset \Gamma$. However, if $\xi(D) \neq \Gamma$, then $\xi(D)^C$ is connected and by continuity of $\xi^{-1}$, $\xi^{-1}(\xi(D)^C) = \xi^{-1}(\xi(D^C)) = D^C$ is connected, which is a contradiction.
If we let $\phi(r,\theta) = i\,\mathbb{arg} (-(2r+(1+r^2)\cos \theta) - i (1-r^2) \sin \theta)$, then we can write $w_1(r,\theta) = \frac{1}{1- r^2}+\frac{r}{|1- r^2|}e^{i \phi(r,\theta)}$.
Now to return to the general problem. As above, we see that $w_0(r,\theta) = \frac{1}{r}e^{-i\theta}$, and for $z_0 = |z_0| e^{i \zeta} \neq 0$, we have $w_{z_0}(r, \theta) = \frac{1}{z_0} w_1(\frac{r}{|z_0|}, \theta-\zeta) $. Hence if $r=|z_0|$, we have the line $w_{z_0}(|z_0|, \theta) = \frac{1}{2 z_0} (1 - i \frac{\sin (\theta-\zeta)}{1+\cos (\theta-\zeta) })$. Finally, if $r \neq |z_0|$, we have $w_{z_0}(r, \theta) = \frac{1}{z_0}\frac{1}{1- (\frac{r}{|z_0|})^2}+\frac{\frac{r}{|z_0|}}{|z_0||1- (\frac{r}{|z_0|})^2|}e^{i (\phi(\frac{r}{|z_0|},\theta)-\zeta)}$.