The parameterization of the circle in rectangular coordinates is given by the the following functions
$$ y = g_1(x) = \sqrt{(1-x^2)} \\ y = g_2(x) = -\sqrt{(1-x^2)} \\ x = g_3(y) = \sqrt{(1-y^2)} \\ x = g_4(y) = -\sqrt{(1-y^2)} $$
for $-1<x<1$ and $-1<y<1$. However the final two (or the first two) are only required because the interval where $x$ (or $y$) is defined is open. Why does that interval need to be open?
Parametrizations are usually taken to be regular, i.e., continuously-differentiable with non-zero derivative.
The notion of differentiability at endpoints is a dicey matter in general, but for the functions $x \mapsto \pm\sqrt{1 - x^{2}}$ the problem is blunt: There's no differentiable extension to $[-1, 1]$, since the graphs have vertical tangents at the endpoints.