I am having trouble compute the image of latitude and logitude lines under the Cayley transform $z \mapsto \frac{z-1}{z+1}$.
So a horizontal line might be $\mathrm{Re}(z) = k \in \mathbb{R}$, then the image curve should be: $\mathrm{Re}[\frac{z-1}{z+1}] = k$. I reasons:
$$ \mathrm{Re}[1 - \frac{2}{z+1}] = k \quad\text{or}\quad \mathrm{Re}[ \frac{1}{z+1}] = \frac{1-k}{2} $$
Maybe it's easier if I just say $w = z+1$. Then some tedious algebra
$$ \frac{1}{w} + \frac{1}{\overline{w}} = \frac{w + \overline{w}}{|w|^2} = \frac{1-k}{2} $$
If we group all the terms to one side we can identify the circle:
$$ |w|^2 + \frac{2}{1-k}(w + \overline{w}) = \left|w + \frac{1}{1-k}\right|^2 - \frac{1}{(1-k)^2} = 0 $$
The center and the radius is as follows:
$$ \left| z + 1+ \frac{1}{1-k}\right| = \frac{1}{1-k} $$
I believe that $\mathrm{Im}(z) = k$ is similar. This answer has some strange behavior as $k \to 1$.
You might want to check that$$\left|\frac{k+it-1}{k+it+1} - \frac{k}{k+1}\right| = \frac 1 {|k+1|}.$$Thus, your transform sends the line $\{z : \operatorname{Re}(z) = k\}$ to a circle with center $\frac{k}{k+1}$ and radius $\frac 1 {k+1}$.