Mapping circles via inversion in the complex plane Consider two unshaded circles $C_r$ and $C_s$ with radii $r>s$ that touch at the origin of the complex plane. The shaded circles $C_1,C_2...C_7$ (labeled in counterclockwise direction sequentially) all touch $C_r$ internally and $C_s$ externally. $C_1$ also touches the real axis and $C_i$ and $C_{ i+1}$ touch for $i=1...6.$
Let $r_i$ denote the radius of $C_i$. Then show that for $i=1,2...,$ 
$$r_i^{-1} + 3r_{i+2}^{-1} = 3r_{i+1}^{-1} + r_{i+3}^{-1}$$
A picture is attached for clarity.

Attempt:
Under inversion, $C_s$ and $C_r$ are mapped to lines and $C_i, i \in \left\{1,...,6\right\}$ are mapped to circles. The map $f(z) = 1/z$ is conformal, so the angles are preserved wherever $f'(z) \neq 0$. Each circle has $4$ tangent points, (except $C_7$) so under the transformation these $\pi/2$ angles are preserved.
The resulting image I have after the transformation is that the $C_i$ are mapped to circles that lie within the two lines and touch the sides.   The only way I see just now to preserve the angles is to have all the circles of the same radii. If $s$ is the radius of $C_s$ and $r$ the radius of $C_r$ then the centre between these two lines is $\frac{1}{2}\left(1/s - 1/2r\right) $. (But I do not think this makes sense.)
Many thanks
 A: 
I'm trying to get an idea of the resultant picture after inversion. Is it something like this?
A: 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.
