How do you know that $y= \sqrt{1-x^2}$ has the shape of a semicircumference?

Question

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.

Answer 1

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).

Answer 2

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:

1. The Euclidean distance $d_2$ from the origin to any point on the semi-circle is constant.
2. This constant, a.k.a radius, is $1$. (Unit circle)
3. 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 $y=\sqrt{1-x^2}$.

Answer 3

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 $(0,0)$.