Find the differential equation of all circles of radius 1 centered on the y-axis I need to find the differential equation of all circles of the form:
$$ x^2 + (y -C_1)^2 = 1$$
Differentiating w.r.t $x$ once yields:
$$ x + (y-C_1) y' =0 $$
Twice:
$$ 1+ (y-C_1) y'' +(y')^2 =0 $$ 
How would I Then find the resulting equation
 A: A hint: 
Eliminate $C_1$ from the first and the second equation.
Note that exactly two circles pass through each point in the strip $-1<x<1$. Therefore we should expect a differential equation of the form $y'^2=\ldots\ $. A second order differential equation has infinitely many solution curves through each point $(x_0,y_0)$ in its domain.
A: Thanks to Christian Blatter.
Starting with
$$x^2 + (y-c)^2 = 1,$$
we obtain 
$$(y-c)y' = -x$$
by differentiation. Then, we square both sides to get
$$(y-c)^2(y')^2=x^2.$$
Using the first equation, we get
$$(1-x^2)(y')^2 = x^2.$$
A: Differentiate given circle equation
$$x^2 + (y-c)^2 = 1 \tag 1$$
to obtain
$$(y-c)y' = -x \tag 2 $$
Differentiate again
$$ \frac{y^{''}}{1+y^{'2}}= \frac{-1}{y-c} \tag3$$
If you want to numerically find a solution, three initial values
$$ (x_i<1,  \;y_i, \;y_i'=\pm \frac{x_i}{y_i-c}) \tag 4 $$
Two slopes are needed for same initial point $x_i$ because by eliminating $(y-c) $ from (1),(2) we get two solutions for each circle in a column of circles centered on y-axis :
$$ y_i'=\pm \frac{x_i}{\sqrt{1-x_i^2}}. \tag 5$$
