Constant curvature I want to proof the following statement and I don't know if I've done correctly

Let $\alpha : [a,b] \to \mathbb{R}^2$ be a curve parametrized by the arc length. Prove that $\alpha$ is a line segment or an arc of a circle iff the curvature is constant.

$\Rightarrow )$ Taking the segment $\alpha: [a,b] \to \mathbb{R}^2$, $\alpha(s)=(a+us, b+vt)$ (with $u^2+v^2=1$) and the arc of the circle $\beta: [a,b] \to \mathbb{R}^2$, $\beta(s)=r(\cos\left(\frac{s}{r}\right), (\sin\left(\frac{s}{r}\right))$ (with $r>0$), is easy to check that $||\alpha ''(s)||=0$ and $||\beta ''(s)||=1/r$ for all $s$. As the curvature is define as $\kappa_\alpha (s)=||\alpha ''(s)||$, we're done. 
$\Leftarrow )$ As the curve $\alpha$ is parametrized by the arc length, we know that $||\alpha '(s)||=1$ for all $s$, so we can write
$$\alpha'(s)=(\cos\theta(s), \sin\theta(s)),$$
where $\theta(s)$ is a function depending on $s$. Applying the first Frenet equation, we have that
$$\kappa_\alpha(s)=\theta'(s).$$ 
If the curvature is constant, then $\theta'(s)$ is too, so $\theta(s)=cs+d$ for some constants $c,d$. If $c=0$, we have a segment line (because then $\alpha(s)=(s\cos c,s\sin c ))$. In other cases, we have an arc of a circle (I don't know exactly how to check it, but I think I have to apply a change of variables).

You did really good. The case for when $c \not = 0$ we have $\alpha'(s) = (\cos(cs+d),\sin(cs+d))$.Let $u(s)=u = cs+d$ then $\alpha'(s)= g(u) = (\cos(u), \sin(u))$ where $g(u):= (\alpha \circ s)(u)$. Observe that the map $u: I:=[0,2\pi] \to u([0,2\pi]):= \tilde{I}$ defined by $s \mapsto cs+d$ is a diffeomorphism and so $g \circ u $ is a reparametrization of a circle. Moreover, we can integrate $g$ to recover $\alpha$. From this we see that $\alpha(s)$ traces a circle and we are done. 
