How do you know that $y= \sqrt{1-x^2}$ has the shape of a semicircumference? A way to explain this is probably that when you square them and then reagroup terms you'll end up having that  $y^2+x^2 =1$ and that forms and circumference, but since $y$ can't be negative it becomes a semicircumference. In other words, what I am looking for is how to explain that $y= \sqrt{1-x^2}$ has the shape of a semicircumference.
 A: Because $x^2+y^2=1$ is known as a circle and $y$ is nonnegative.
(Yes: exactly as you think, modulo the fact that a circumference is a number).
A: If you want to work out where the equation comes from, which is what your post seems to be requesting, you can derive it as follows:
A unit semi-circle in the upper plane centered about the origin satisfies three properties:


*

*The Euclidean distance $d_2$ from the origin to any point on the semi-circle is constant.

*This constant, a.k.a radius, is $1$. (Unit circle)

*The $y$ values must be non-negative. (Upper plane)


The Euclidean distance between $(x,y)$ on the semi-circle and the origin is therefore
\begin{align}
d\big((x,y),(0,0)\big)_2 &= \sqrt{(x-0)^2+(y-0)^2}
= \sqrt{x^2+y^2} = 1\\
\implies\quad x^2+y^2&=1^2 = 1
\end{align}
This has two sets of solutions $y = \pm \sqrt{1-x^2}$. Ensuring 3 we get the only solutions $y=\sqrt{1-x^2}$.
A: First of all, writing $y = \sqrt{1-x^2}$ makes sense only if $-1 \le x \le 1$ and $y \ge 0$. This means that, when you plot the graph of this function, you should look only at the sub-region of $\mathbb{R}^2$ given by $-1 \le x \le 1$ and $y \ge 0$. 
Now observe that if you square both sides, you get $x^2+y^2=1$, which is the equation of a circle centered at $(0,0)$, with radius $1$. If you are not convinced by this, you can see it in polar coordinates: set $x = r\cos\theta, y = r\sin \theta$, with $r > 0$ and $\theta \in [0,2\pi)$. So $$1 = x^2+y^2 = r^2(\cos^2\theta+\sin^2\theta) = r^2$$
and since $r > 0$ you end up with $r = 1$. This parametrization is centered at the origin, so it gives you all the points whose distance from the origin is $1$, hence a circle of radius $1$. 
To sum up, the set of points $(x,y)$ such that $y = \sqrt{1-x^2}$ is a subset of the circle. In particular, $y \ge 0$, so it is the upper half of the circle of radius $1$ centered at $(0,0)$.
