I’m trying to follow the following proof: Wikipedia link
It starts with: Let $P:\vec p$ be a point, $c: \vec x^2-r^2=0$ a circle with the origin as its center.
And then shows a picture of a circle.
How is this equation $c: \vec x^2-r^2=0$ a circle? Doesn’t it only specify $x^2 = r^2$? I was expecting a circle to be defined using $x^2 + y^2 = r^2$.
Note the equation is $\vec{x}^2−r^2=0$, where $\vec{x}$ is a vector, not a scalar like $x$, so it doesn't specify that $x^2=r^2$. Also, the $\vec{x}^2$ term means the dot product of the vector with itself, i.e., $\vec{x}\cdot\vec{x}$. Thus, in Cartesian co-ordinates, if $\vec{x}=(x,y)$ (where the $x$ in $(x,y)$ is a scalar), we then get
$$(x,y)⋅(x,y)−r^2=0 \;\;\to\;\; (x^2+y^2)−r^2=0 \;\;\to\;\; x^2+y^2=r^2$$
which is the equation of a circle, of radius $r$, centered at the origin.