curvature of a circle

Let $C_{r}$ be a circle of radius $r$ centered at the origin.

A canonical parameterization of the curve is (counterclockwise)

g(s)=r\\left(\\cos\\left(\\frac{s}{r}\\right),\\sin\\left(\\frac{s}{r}\\right)\\right)

for $s\\in(0,2\\pi r)$ (actually this leaves out the point $(r,0)$ but this could be treated via another parameterization taking $s\\in(-\\pi r,\\pi r)$ )

Differentiating the parameterization we get

\\mathbf{T}=g^{\\prime}(s)=\\left(-\\sin\\left(\\frac{s}{r}\\right),\\cos\\left(\\frac{s%}{r}\\right)\\right)

and this results in the normal

\\mathbf{N}=J\\cdot\\mathbf{T}=-\\left(\\cos\\left(\\frac{s}{r}\\right),\\;\\sin\\left(%\\frac{s}{r}\\right)\\right)=-\\frac{g(s)}{r}

Differentiating $g$ a second time we can calculate the curvature

\\mathbf{T}^{\\prime}=-\\frac{1}{r}\\left(\\cos\\left(\\frac{s}{r}\\right),\\;\\sin\\left%(\\frac{s}{r}\\right)\\right)=\\frac{1}{r}\\mathbf{N}

and by definition

\\mathbf{T}^{\\prime}=k\\mathbf{N}\\;\\;\\therefore\\;k=\\frac{1}{r}

and thus the curvature of a circle of radius $r$ is $\\displaystyle{\\frac{1}{r}}$ provided that the positive direction on the circle is anticlockwise; otherwise it is $\\displaystyle{-\\frac{1}{r}}$ .