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.


1 Answer 1


Use the formula (see here) for the radius of curvature :


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).

  • $\begingroup$ Any comment, two days later ? $\endgroup$
    – Jean Marie
    Dec 12, 2023 at 23:04

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