Differential equation for all circles in a plane.

Question

We want the differential equation for all circles in a plane.

$$(x-h)^2+(y-k)^2=r^2$$

I have done the following $$(x-h)+(y-k)y'=0 \\ 1+(y')^2+yy''-ky''=0 \\ \frac{1+(y')^2+yy''}{y''}=k \\ \frac{(2y'y''+(y'')^2+yy''')y''-(1+(y')^2+yy'')y'''}{(y'')^2} = 0 \\ y'''+(y')^2y'''-(y'')^2-2y'y''=0$$

Is this solution correct? I am not sure if cancelling $(y'')^2$ is the correct.

Answer 1

$$\frac{1+(y')^2+yy''}{y''}=k \\ \frac{(2y'y''+\color{red}{(y'')^2}+yy''')y''-(1+(y')^2+yy'')y'''}{(y'')^2} = 0 $$

I think the part in red should be $\color{blue}{y'y''}$ instead of $\color{red}{(y'')^2}$.

Alternatively, picking up at:

$$ 1+(y')^2+yy''-ky''=0 \tag{$*$}$$

Differentiating again: $$3y'y''+yy'''-ky'''=0 \iff k=\frac{3y'y''}{y'''}+y$$ Substitution into $(*)$: $$1+(y')^2+yy''-\left(\frac{3y'y''}{y'''}+y\right)y''=0$$ Multiplying by $y'''$ and simplifying: $$\left(1+(y')^2\right)y'''-3y'(y'')^2=0$$