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
when we move from $(0,0) \rightarrow (1,0)$ then $0\leq x\leq 1$ and $y=0$ under the given transformation $w=\dfrac{1}{z+1}=\dfrac{x+1-iy}{(x+1)^2+y^2}$
then $0\leq x\leq 1\implies 1\leq x+1\leq 2\implies 1\geq\dfrac{1}{x+1}\geq \dfrac{1}{2}$ Thus under the transformation if $w=u+iv$ then $\dfrac{1}{2}\leq u\leq 1$ and $v=0$
again when $0\leq y\leq 1$
$\dfrac{1}{1+y^2}-i\dfrac{y}{1+y^2}$ provided $x=0$
Hence the real part $0\geq \frac{-y}{1+y^2}\geq \dfrac{-1}{2}$
and the imaginary part $\dfrac{1}{2}\leq \dfrac{1}{1+y^2}\leq 1$.Now can you find the rest