Show that the curve defined by

$\gamma (s):=(a\mathrm{cos}\left(\frac{s}{a}\right),a\mathrm{sin}\left(\frac{s}{a}\right))$

Is on a circle with the radius $a$ and the center (0,0),also show that $\gamma (s)$ has been parameterized by its arc length.

I know that the parametric equation of $a$ circle with radius a and center $({x}_{0},{y}_{0})$ is :

$x={x}_{0}+a\mathrm{cos}\left(t\right)$

$y={y}_{0}+a\mathrm{sin}\left(t\right)$

If we denote the parametric equation of a circle with $\gamma (t)=({x}_{0}+a\mathrm{cos}\left(t\right),{y}_{0}+a\mathrm{sin}\left(t\right))$,then we have:

$\frac{d\gamma (t)}{dt}=(-a\mathrm{sin}\left(t\right),a\mathrm{cos}\left(t\right))$

$\Vert \frac{d\gamma (t)}{dt}\Vert =\sqrt{{a}^{2}}=a$

So the arc length is :

$s={\int}_{0}^{t}\Vert \frac{d\gamma (\tau )}{dt}\Vert d\tau =a{\int}_{0}^{t}\phantom{\rule{thickmathspace}{0ex}}d\tau $

Which implies $s=at$.

And if we parameterize the circle by its arc length then:

$\gamma (s)=(a\mathrm{cos}\left(\frac{s}{a}\right),a\mathrm{sin}\left(\frac{s}{a}\right))$

Which is the given parametrization.

But I have not shown that the center is at the origin.