Find the equation of this bijection from $ \mathbb{R} $ to $ (0,1) $. I need some help (hints or an answer) in finding the actual equation of this bijections from the reals $ \mathbb{R} $ to $ (0,1) $. We may assume that the radius of the circle is $ \dfrac{1}{2} $.

Also, is it possible to transform this geometric representation to an order-preserving isomorphism?
Thanks!
Also, I am aware there are much simpler isomorphisms between $ \mathbb{R} $ and $ (0,1) $.
 A: Just do the work.
Assume that the circle is $x^2+y^2 = 1$. Assume that the line is $y=-k$ for some positive real number $k$.
Then, the point $(x, k)$ will get mapped to the point $(\frac {x}{\sqrt{x^2+k^2}}, \frac {-k} { \sqrt{x^2+k^2}})$.
If you want to map it to $(0, 1)$ as opposed to $(-1, 1)$, then take the transformation $t \rightarrow \frac {t+1} {2} $. Alternatively, use another circle but that makes the derivations ugly.
A: The point $x$ in $(0,1)$ is sent to the point $y(x)$ in $\mathbb R$ if there exists some angle $\theta$ in $(0,\pi)$ such that $\cos\theta=2x-1$ and $y=a\cot\theta$, where $a$ is the distance between the center of the circle and the line. Since $\sin\theta=\sqrt{1-\cos^2\theta}$, this yields
$$
y(x)=\frac{a(2x-1)}{\sqrt{1-(2x-1)^2}}=\frac{a(2x-1)}{\sqrt{4x(1-x)}}.
$$
For $y\mapsto x(y)$ sending $\mathbb R$ to $(0,1)$, use the inverse mapping
$$
x(y)=\frac12\left(1+\frac{y}{\sqrt{y^2+a^2}}\right).
$$
Both functions $x\mapsto y(x)$ and $y\mapsto x(y)$ are increasing diffeomorphisms.
