Show that the curve defined by $$\gamma(s):=\left(a\cos\left(\frac{s}{a}\right),a\sin\left(\frac{s}{a}\right)\right)$$
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 : $$x=x_{0}+a\cos\left(t\right)$$ $$y=y_{0}+a\sin\left(t\right)$$ If we denote the parametric equation of a circle with $\gamma(t)=\left(x_{0}+a\cos\left(t\right),y_{0}+a\sin\left(t\right)\right)$,then we have:
$$\frac{d\gamma(t)}{dt}=\left(-a\sin\left(t\right),a\cos\left(t\right)\right)$$
$$\left\Vert \frac{d\gamma(t)}{dt}\right\Vert=\sqrt{a^{2}}=a$$
So the arc length is :
$$s=\int_{0}^{t}\left\Vert \frac{d\gamma(τ)}{dt}\right\Vert dτ=a\int_{0}^{t} \;dτ$$
Which implies $s=at$.
And if we parameterize the circle by its arc length then:
$$\gamma(s)=\left(a\cos\left(\frac{s}{a}\right),a\sin\left(\frac{s}{a}\right)\right)$$
Which is the given parametrization.
But I have not shown that the center is at the origin.
You are given the parametrization in the form $x(s) = a\cos\left(\frac s a\right) = 0 + a\cos\left(\frac s a\right)$, $y(s) = a\sin\left(\frac s a\right)=0 + a\sin\left(\frac s a\right)$. According to what you say you know about parametrizing circles, that $x(t) = x_0 + a\cos(t)$ and $y(t) = y_0 + a\sin(t)$ parametrizes a circle centered at $(x_0,y_0)$, then you should be able to conclude that $s\mapsto \big(x(s),y(s)\big)$ parametrizes a circle centered at $(x_0,y_0) = (0,0)$ of radius $a$.