Show that a circle of radius $r$ has radius of curvature equal to $r$

Question

There is this statement from the book Differential Geometry of Curves and Surfaces by Do Carmo that I am trying to verify it myself.

The inverse $R=1/k$ of the curvature is called the radius of curvature at $s$. Of course, a circle of radius $r$ has radius of curvature equal to $r$, as one can easily verify.

So, I am trying to show that a circle of radius $r$ has radius of curvature equal to $r$:

Let $\alpha(s)=(r\text{ }\cos s, r\text{ }\sin s)$, for $r\geq0$ and $s$ is the arc length as the curve parametrisation.
Then $\alpha''(s)=(-r\text{ }\cos s, -r\text{ }\sin s)$, and $k=|\alpha''(s)|=r$.
So $R=1/k=1/r$.

Then how are we going to show that $R=r$?

I doubt that the statement should be a circle of radius $r$ has radius of curvature equal to 1? Am I correct?

Thanks for the help and clarification!

Answer 1

Thanks Daniel Fischer for the helpful comments.

The curve $\alpha(s)=(r\text{ }\cos s, r\text{ }\sin s)$ is not parametrised by arc length since $|\alpha'(s)|\neq1$

Hence we define a new curve $\beta(s)=\alpha(1/r\cdot s)$ for $r>0$. Then $\beta(s)=(r\text{ }\cos (1/r\cdot s), r\text{ }\sin (1/r\cdot s))$ and $\beta'(s)=(-\sin(1/r\cdot s), \cos(1/r\cdot s))$. So $|\beta'(s)|=1$. Furthermore, $\beta''(s)=(-1/r\text{ }\cos (1/r\cdot s), -1/r\text{ }\sin (1/r\cdot s))$, so $k=|\beta''(s)|=1/r$.

Hence $R=1/k=1/(1/r)=r$.