# Differential equation whose solution are circles

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?

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

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
$$(y')^6 + 3 (y')^4 - r^2 (y'')^2 + 3 (y')^2 + 1 = 0$$
• @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) Dec 11, 2023 at 7:03