I know $r^2 = (x-h)^2 + (y-k)^2$ (A)
Differentiate (A)
$0 = 2(x-h) + 2(y-k)^2y'$
$0 = x - h + (y-k)^2y'$ (B)
Differentiate B
$0 = 1 + 2(y-k)y'^2 + (y-k)^2y''$ (C)
At this point I can solve for h, solving for k I think I would have to expand the terms in (C). I am unable to get the answer: $[1 + y'^2]^3 = r^2(y'')^2$
Any help would be appreciated.
Use the formula (see here) for the radius of curvature :
$$R=\frac1r=\frac{(1+y'^2)^{3/2}}{y''}$$
Why is it necessary to square this relationship to have the final answer ? In order to take into account the upper and the lower part of the circles ( a circle has two cartesian equations).