Find the differential equation of all circles of radius a 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. 
 A: 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
$$
((y')^2+1)^3=(ay'')^2
$$
A: 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.
A: 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
$$  \frac{1}{\sqrt{ r^2 +r'^2  }} + \frac{r'^2 - r r ^{''}}{ r^2 +r'^2  } \frac{1}{\sqrt{ r^2 +r'^2  }} \tag{5=1}  $$
