if the curvature is constant and positive, then it is on the circunference I'm trying to prove that if $\alpha(t)=(x(t),y(t))$ is a $C^2$ regular curve $(\alpha'\neq0)$ with constant and positive curvature, then $\alpha$ is on the circunference and if $\alpha$ is the circunference, then its curvature is constant.
I solved the second part since $\alpha$ is the circunference, then $\alpha(t)=(r\cos\theta,r\sin\theta)$, so $|\alpha'(t)|=r$, I need help in the first part.
Thanks a lot.
 A: Here are my hints, following the argument given by John Oprea in his text "Differential Geometry and Its Applications". If you are still confused, Oprea's text has a complete proof of this problem, but try and see if you can solve it yourself first given these hints.
Suppose, without loss of generality, that $\alpha$ is a unit speed curve.
Suppose $\alpha$ is part of a circle. Then since every point on $\alpha$ is a fixed distance $r$ away from some point $p$, we can write:
$(\alpha(t) - p)\cdot(\alpha(t) - p) = r^{2}$
Differentiate this expression several times with respect to $t$. See if you can't use the expressions you obtain to show that $\kappa \neq 0$, i.e., $\kappa > 0$. Then see if you can't get an expression involving $\frac{d\kappa}{ds}$ and use this expression to conclude $\frac{d\kappa}{ds}$ is 0, i.e., $\kappa$ is constant. 
Now suppose $\alpha$ has positive constant curvature. See if you can't show that the curve $\gamma(t) = \alpha(t) + \frac{1}{\kappa}N$ is actually constant, i.e., a point $p$. From here, see if you can't use this to show that $\alpha$ is a fixed distance away from from $p$ using what you know about $N$. 
