Mapping of Disk to sphere homeomorphically, radius of latitude and annulus 
I have some questions related to the example in the above attached image. The example came from a text titled: Elements of Topology by Tej Bahadur Singh 
In the example, Sinh showed how $\mathbb{D}^{2}/\mathbb{S}^{1}$ is homeomorphic to $\mathbb{S}^{2}\\$
Let the map $h:x\rightarrow\frac{x}{1-||x||}$ be a homeomorphism of $\mathring{(\mathbb{D}^{2})}$ onto $\mathbb{R}^2$.  Let $p=(0,0,1)\in\mathbb{S}^2$, then the map $g:\mathbb{S}^2-\{p\}\rightarrow\mathbb{R}^2,$ $(x_0,x_1,x_2)\mapsto\frac{1}{1-x_2}(x_0,x_1), $is a homeomorphism.  The map $g^{-1}h$ is a homeomorphism between $\mathring{(\mathbb{D}^{2})}$ and $\mathbb{S}^2-\{p\}.$  He then defines a function $f:\mathbb{D}^{2}\rightarrow\mathbb{S}^{2}$ by
$f(x) = \left\{
        \begin{array}{ll}
            g^{-1}h(x) & \quad  \text{for $x\in \mathbb{D}^2-\mathbb{S}^1$ and}\\
            p & \quad \text{for $x\in \mathbb{S}^1$}
        \end{array}
    \right.\\$
Singh states that the function $f$ maps a concentric circle of radius $r$ in $\mathbb{D}^2$ homeomorphically onto a parallel at the latitude $\frac{2r-1}{1-2r+2r^2}$ in $\mathbb{S}^2$ and takes the radii in $\mathbb{D}^2$ onto the meridians running from the south pole to the north pole.
To show the continuiuty for the case of $x\in\mathbb{S}^1$, pick a small ball $B(p;\epsilon)\subset U$, then $V=f^{-1}(B(p;\epsilon))$ is the annulus $\{x\in\mathbb{D}^2|r < ||x|| \leq 1\}, \text{where } r^2=\frac{4-\epsilon^2}{4+2\epsilon\sqrt{4-\epsilon^2}}$
The two things I don't understand are how Singh got the expression:
$\frac{2r-1}{1-2r+2r^2}$ for the function $f$ and also the radius of the annulus: $r^2=\frac{4-\epsilon^2}{4+2\epsilon\sqrt{4-\epsilon^2}}$
I feel like I am missing something, because usually, this example is done by taking the inverse map of stereographic projection which has been asked about quite a few times on this community. If anyone can help me to shed some light no how Sinh got his expression, it will be much appreciated.  Thank you in advance.
 A: I'll answer the first question only as I don't have more time right now.
The key to the first question is that latitude means $z$-value here. In other words, it should be really altitude. You can guess it because the range of the function $\frac{2r-1}{1-2r+2r^2}$ is $[-1,1]$. Now here is how to see the formula:
First, $h$ maps the circle of radius $r$ onto the circle of radius $\color{red}{\frac r{1-r}}$.
Second, $g$ maps the circle of altitude $x_2$ onto the circle of radius $\|\frac 1{1-x_2}(x_0,x_1)\|=\frac 1{|1-x_2|}\sqrt{x_0^2+x_1^2}=\frac{\sqrt{1-x_2^2}}{|1-x_2|}=\color{green}{\frac{\sqrt{1+x_2}}{\sqrt{1-x_2}}}$.
Now setting $\color{red}{\frac r{1-r}}=\color{green}{\frac{\sqrt{1+x_2}}{\sqrt{1-x_2}}}$, the game is to see why $x_2=\frac{2r-1}{1-2r+2r^2}$. I'll drop the $_2$ from now on.
$$\begin{eqnarray}\frac {r^2}{(1-r)^2}&=&\frac{1+x}{1-x}\\(1-x)\frac {r^2}{(1-r)^2}&=&{1+x}\\ x\cdot\left(1+\frac {r^2}{(1-r)^2}\right)&=&\frac {r^2}{(1-r)^2}-1\\x&=&\frac{r^2-(1-r)^2}{r^2+(1-r)^2}\\ x&=&\frac{2r-1}{1-2r+2r^2}\end{eqnarray}$$
