Find the differential equation of all circles of radius a [closed]

Can someone please post a detailed step-by-step procedure. Given the circle with a radius a, what is the differential equation of the circle.

Circle equation $$ (x-C_1)^2+(y-C_2)^2=a^2\quad (0) $$ Differentiate twice by $x$ $$ (x-C_1)+y'(y-C_2)=0\quad (1) $$ $$ 1+(y-C_2)y''+(y')^2=0\quad (2) $$ From $(2)$ we obtain $$ C_2=y+\frac{(y')^2+1}{y''} $$ Then substitute in $(1)$ and $(0)$ $$ (x-C_1)-y'\frac{(y')^2+1}{y''}=0\quad(3) $$ $$ (x-C_1)^2+\left(\frac{(y')^2+1}{y''}\right)^2=a^2\quad(4) $$ From $(3)$ we obtain $$ x-C_1=y'\frac{(y')^2+1}{y''} $$ Then substitute in $(4)$ $$ \left(y'\frac{(y')^2+1}{y''}\right)^2+\left(\frac{(y')^2+1}{y''}\right)^2=a^2 $$ After some simplifications we get $$ ((y')^2+1)^3=(ay'')^2 $$

From the implicit equation of the circle $(x-u)^2+(y-v)^2=a^2$, you get $$x'(x-u)+y'(y-v)=0$$ by implicit differentiation. Add the initial condition $$x(0)=u+a, \quad y(0)=v$$

You can write the differential equations as $$ x'=-y+v, \quad y' = x-u $$ which is especially nice for circles centered at the origin.

We can express standard result/formula for curvature rectangular coordinates

$$ \frac{y''}{(1+y'^2)^{3/2}} =\frac{1}{a} $$

as the required differential equation.

EDIT 1:

In polar co-ordinates the ODE for curvature is

$$ \dfrac { r^2 + 2 r ^{'2 }- r r ^{"} }{ (r^2 +r'^2)^{3/2}}=\frac{1}{a} \tag{1}$$

The above can be derived from intrinsic/natural differential equation of a circle is

$$ \frac{d \phi }{ds} = \frac{d (\theta + \psi ) }{ds}=\frac{1}{a}\tag{2}$$

$$ =\dfrac{ \sin \psi}{r} + \frac{d}{ds} ( \tan ^{-1} \frac{r}{r'} ) \tag{3} $$

where $\phi$ is angle to x-axis, $ \psi$ is between arc and radius vector, $$ \tan \psi = \dfrac {r}{r^{'}} \tag{4} $$

Introducing above into (3) and differentiating, LHS is

$$ \frac{1}{\sqrt{ r^2 +r'^2 }} + \frac{r'^2 - r r ^{''}}{ r^2 +r'^2 } \frac{1}{\sqrt{ r^2 +r'^2 }} \tag{5=1} $$