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

$$ \mathbf{T} = g'(s) = \left( -\sin \left( \frac{s}{r} \right), \cos \left( \frac{s}{r}\right) \right)$$

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

$$\mathbf{T}' = -\frac{1}{r}\left(\cos\left(\frac{s}{r}\right),\;\sin\left(\frac{s}{r}\right)\right) = \frac{1}{r} \mathbf{N}$$

and by definition

$$ \mathbf{T}' = k\mathbf{N}\;\; \therefore\; k = \frac{1}{r} $$

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 $\displaystyle{-\frac{1}{r}}$