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I want to find a differential equation, which has as solutions circles of radius $r$. And the equation needs to be of the form $F(y'',y',y)=0$.

The equation $$(x-x_0)^2 +(y-y_0)^2 = r$$

describes a circle of radius $r$ centered at the point $(x_0,y_0)$. If we differentiate this relation twice with respect to $x$, we have the equation $$2+2y''(y-y_0)+2(y')^2=0$$

which is in the correct form. I think this is too simple to be correct. Any suggestions?

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    $\begingroup$ You want $r^2$, not $r$, on the right side of the equation for a circle. $\endgroup$ Feb 24, 2021 at 19:33

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Your differential equation is satisfied by any circle of any radius whose centre has $y$ coordinate $y_0$. I would interpret the question as saying that you want $x_0$ and $y_0$ to be arbitrary but $r$ to be fixed. So you want to take your original equation and its first and second derivatives and eliminate $x_0$ and $y_0$. I get

$$ (y')^6 + 3 (y')^4 - r^2 (y'')^2 + 3 (y')^2 + 1 = 0$$

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  • $\begingroup$ Is the logic behind this, that we may invert the process. That is, we start with this differential equation and end up with out original relation for circles? $\endgroup$ Feb 25, 2021 at 13:27
  • $\begingroup$ @Robert Israel The equation you obtain can be written $((y')^2+1)^3=r^2(y'')^2$ with a clear connection with the formula giving the radius of curvature (see my recent answer here) $\endgroup$
    – Jean Marie
    Dec 11, 2023 at 7:03

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