Find the differential equation of all circles of radius 1 centered on the y-axis

Question

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

Answer 1

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.

Answer 2

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

$$(1-x^2)(y')^2 = x^2.$$

Answer 3

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 :

$$ y_i'=\pm \frac{x_i}{\sqrt{1-x_i^2}}. \tag 5$$