In what the parametrisation $(x,\sqrt{1-x^2})$ of the circle is interesting? We all know the parametrisation $\gamma (t)=(\cos(t),\sin(t))$ of the circle that has the advantage to be smooth and is easy to use. Sometime, my teacher use the parametrisation $$\left\{\varphi(t)=\left(t,\sqrt{1-t^2}\right)\mid t\in [-1,1]\right\}\cup\left\{\tilde \varphi(t)=\left(t,-\sqrt{1-t^2}\right)\mid t\in [-1,1]\right\},$$
and I would be curious to know in what this parametrization can be interesting. Moreover, the speed goes to infinity when $x\to \pm 1$, so it's not very good, is it ? I would be curious to know is what this paramtrisation is more interesting rathe $t\longmapsto (\cos t,\sin t)$.
 A: The one you are talking about is often used as an example for changing charts. With that parametrization you cannot parametrize the whole circumference, because at $x= \pm 1$ you have problems. Instead, you parametrize the upper and lower semicircunference, except two intervals containing $x= \pm 1$. Then, you parametrize those points by doing the projection on the $y$ axis. Finally, you want to show that in those regions where you can use both parametrizations, you can use both. This is called transition map (if I am not mistaken) and it is needed in order to check that the whole parametrization of your curve (called atlas) is coherent. 
I am not very used to this kinda stuff, these should be the principal ideas behind it, though
A: I've found it surprisingly useful.  Below are two example cases that come to mind.

If you wanted to calculate the area of a circular disk between two parallel lines, pick a coordinate system with $x$ axis perpendicular to those lines, and origin at the center of the disk. Let $x_1 \le x_2$ be the $x$ coordinates of the lines, and $r \gt 0$ the radius of the disk. Then,
$$\begin{aligned}
A &= 2 \int_{x_1}^{x_2} \sqrt{r^2 - x^2} \, d x \\
\;&= \Biggl\lvert_{x_1}^{x_2} x \sqrt{r^2 - x^2} + r^2\arcsin\left(\frac{x}{r}\right)\Biggr. \\
\;&=  x_2 \sqrt{r^2 - x_2^2} - x_1 \sqrt{r^2 - x_1^2} + r^2\left(\arcsin\left(\frac{x_2}{r}\right) - \arcsin\left(\frac{x_1}{r}\right)\right)
\end{aligned}$$

The formula for an $r$-radius circle centered at $(x_1, 0)$ is
$$y(x) = \pm \sqrt{r^2 - (x - x_1)^2}$$
Let's say you have two spherical bubbles in 3D, touching or intersecting each other. You know their intersection is a circle, perpendicular to the line between the centers of the two spheres. You want to find where the center of the circle is, and what the radius of the intersection circle is.
Pick a 2D coordinate system where one sphere (radius $r_0 \gt 0$) is at origin, and the other is on the $x$ axis, at $(x_1, 0)$, with $x_1 \gt 0$, and radius $r_1 \gt 0$. Point $(x, 0)$ is the center of the circle where the two spheres intersect, if and only if
$$\sqrt{r_0^2 - x^2} = \sqrt{r_1^2 - (x - x_1)^2}$$
Solving for $x$ is easy, because you can start by squaring both sides. (You can do that, because they are both positive and equal, or both zero.)
You get
$$x = \frac{r_0^2 + x_1^2 - r_1^2}{2 x_1}$$
The radius of the intersection circle is obtained by substituting back to the previous equation (left or right side):
$$r = \frac{1}{2}\sqrt{2 r_0^2 + 2 r_1^2 - x_1^2 - \frac{(r_0^2 - r_1^2)^2}{x_1^2}}$$
A: The other parametrisation comes from pythagorean version of circle:
 $ x^2+y^2=1 $
This version of circle is of form $f(x,y)=1$.
To get the parametrisation, you need to use substitution $y=f(x)$, i.e.
$f(x,f(x))=1$ and thus $x^2+f(x)^2=1$, which means $f(x)^2 = 1-x^2$ which gives $f(x)=\pm \sqrt{1 - x^2}$.
The interesting part is that this same process can be done to any equation of the form $f(x,y)=c$.
