Proving Constant Curvature of r(s) is a Circle

[hholzer]

If we parameterize the arc length of a vector valued
function, say, r(s) and r(s) has constant curvature
(not equal to zero), then r(s) is a circle.

Thus, |T'(s)| = K but to prove it we would need
to show |T'(s)| = K => <-Kcos(s), -Ksin(s)>
and integrate component-wise two times,
right?

 

[Tinyboss]

If your function is into R^2...else it could be way more complicated. I'd try to find a point x such that |r(s)-x| is constant.