It appears you have all the information to draw. The image is all points outside of that circle, and below the real axis. If you look at just the image of the boundary, the real axis mapped to the real axis and the horizontal line at $y=1/(2c)$ mapped to a circle, and the image is the region between those two.
Concerning another line mapping to a circle, this is a general property of a class of complex transformations called Linear Fractional Transformations, or Moebius Transformations, see http://en.wikipedia.org/wiki/M%C3%B6bius_transformation
which map lines and circles to other lines and circles (or if you like, just circles to circles, treating lines as circles with the origin at infinity). In this case, your computations show quite explicitly how lines get transformed into circles.
The two circles $C_r$ and $C_s$ are mapped to the parrallel lines $\left\{ \operatorname{Re} w = \frac{1}{2r}\right\}$ and $\left\{ \operatorname{Re} w = \frac{1}{2s}\right\}$ by the inversion. Since the inversion maps circles to circles (where a straight line is counted as a circle of infinite radius), it maps each of the $C_i$ to a circle touching both of these parallel lines. Also, the image of $C_1$ touches the real axis (since $C_1$ does, and the inversion maps the real axis $\cup \{\infty\}$ to itself), and further, each circle $C_k$ for $k > 1$ touches the circles $C_{k-1}$ and $C_{k+1}$, hence so do their images.
The distance between the two parallel lines is $\frac{1}{2s} - \frac{1}{2r}$, so the images of the $C_k$ have the common radius
$$R = \frac{1}{2}\left(\frac{1}{2s} - \frac{1}{2r}\right),$$
and their centres all lie on the line $$\left\{ \operatorname{Re} w = M\right\};\qquad M := \frac{1}{2}\left(\frac{1}{2s} + \frac{1}{2r}\right).$$
It follows that the centre of the image $C_k'$ of $C_k$ is
$$b_k = M - (2k-1)\cdot iR,$$
and hence the defining equation of $C_k'$ is
$$\lvert w-b_k\rvert^2 = R^2,$$
or
$$w\overline{w} - b_k\overline{w} - \overline{b_k}w + (\lvert b_k\rvert^2 - R^2) = 0.$$
Applying the inversion, we see that $C_k$ has the defining equation
$$1 - b_k z - \overline{b_kz} + (\lvert b_k\rvert^2-R^2)z\overline{z} = 0.$$
Since $\lvert b_k\rvert^2 = M^2 + (2k-1)^2R^2 > R^2$, we can divide and obtain the equivalent equation
$$z\overline{z} - \frac{\overline{b_k}}{\lvert b_k\rvert^2-R^2}\overline{z} - \frac{b_k}{\lvert b_k\rvert^2 - R^2} z + \frac{1}{\lvert b_k\rvert^2-R^2} = 0,$$
or
$$\left\lvert z - \frac{\overline{b_k}}{\lvert b_k\rvert^2-R^2} \right\rvert^2 = \left(\frac{R}{\lvert b_k\rvert^2-R^2}\right)^2.$$
The desired relation between the radii should not be difficult to obtain from that.
Best Answer
Linear fractional transformations are determined by their values at three points. Furthermore, lines and circles get transformed to lines and circles.
The trick to problems like this is:
In the input space, you have two key lines and circles. To get the output shape you want:
Note that your two shapes only intersect in one point — so after transformation they will still only intersect in one point. If that point of intersection is $\infty$, you're guaranteed that they are parallel lines.
So, if you ensure the output has this configuration of three points, you're guaranteed that the output of the transformation will be a pair of vertical lines. You just need to play with the orientation (e.g. swap the two points chosen in the second bullet) to toggle whether or not the shaded region gets mapped to the interior or the exterior of the strip.