Differential geometry, show a curve parameterizes a circle So I have this exercise from differential geometry and I am struggling with how to show that this is indeed a valid parameterization.
Let $n(\phi) = (\cos(\phi), \sin(\phi))$. Show that the curve 
$$c(\phi) = n'(\phi) - n(\phi)$$
parametrizes a circle.
What I have so far:
By this definition, $$c(\phi)= (-\cos(\phi)-\sin(\phi),\cos(\phi)-\sin(\phi))$$
so then $$c'(\phi)=(-\cos(\phi) + \sin(\phi),-\sin(\phi)-\cos(\phi))$$
which implies $$|c'(\phi)|=\sqrt[2]{(-\cos(\phi)+\sin(\phi))^2 +(-1^2)(\sin(\phi)+\cos(\phi))^2}=\sqrt[2]{\cos^2(\phi)+\sin^2(\phi) -2\cos(\phi)\sin(\phi)+\sin^2(\phi)+\cos^2(\phi)+2\cos(\phi)\sin(\phi)}=\sqrt[2]{2}\neq 0$$
So since $|c'(\phi)|\neq 0$, then it is smooth. But I don't think this is sufficient for showing that the provided curve parameterizes a circle.  
 A: With
$c(\phi) = (-\cos \phi - \sin \phi, \cos \phi - \sin \phi), \tag 1$
we have
$\vert c(\phi) \vert^2 = (\cos \phi + \sin \phi)^2 + (\cos \phi - \sin \phi)^2$
$= \cos^2 \phi + 2 \cos \phi \sin \phi + \sin^2 \phi + \cos^2 \phi - 2\cos \phi \sin \phi + \sin^2 \phi$
$= 2(\cos^2 \phi + \sin^2 \phi) = 2, \tag 2$
whence
$\vert c(\phi) \vert = \sqrt 2, \tag 3$
which shows that the curve $c(\phi)$ is contained in the circle of radius $\sqrt 2$ centered at the origin.  Our OP geoplanted has shown that also
$\vert c'(\phi) \vert = \sqrt 2 \tag 4$
(in fact it is essentially the same calculation as (2)); thus the curve $c(\phi)$ is regular, and of constant speed $\sqrt 2$; this in turn implies it completely traverses the circle of radius $\sqrt 2$ centered at the origin whenever $\phi$ increases by
$\Delta \phi = \dfrac{2\pi \sqrt 2}{\sqrt 2} = 2\pi; \tag 5$
we thus see that not only does $c(\phi)$ lie in this circle, it in fact passes through each and every one of the circle's points, hence $\text{Range} (c(\phi))$ is said circle of radius $\sqrt 2$ centered at $(0, 0)$.
A: Just note that thank's to the addition formula for $\sin$ and $\cos$ we have : 
$$-\cos \theta - \sin \theta = -\sqrt{2} \sin \left ( \theta + \frac{\pi}{4} \right)$$
$$\cos \theta - \sin \theta = \sqrt{2} \sin \left ( \frac{\pi}{4} - \theta \right)$$
Hence it parameterizes a circle of radius $\sqrt{2}$ centered at $(0,0)$.
