differential equation of family of circles passing through origin How do I find the DE of all circles passing through origin? I tried something like this
The family of circles passing through the origin is given by
     $$
  (x- r \cos \theta )^2 + (y- r\sin \theta)^2 = r^2 
  $$Differentiating once, we get
     $$ 
   2 ( (x- r \cos \theta) + (y - r \sin \theta) y') = 0
  $$
     Differentiating again, we get
     $$
   1 + y'^2  + (y - r \sin \theta) y'' = 0 
  $$
How to get rid of $r$ and $\theta$ algebraically? Is there any other approach?
 A: Let the the circle be of radius $r$. Then,
$$\tag1(x-r\cos\theta)^2+(y-r\sin\theta)^2=r^2$$
$$\tag22(x-r\cos\theta)+2(y-r\sin\theta)y'=0$$
$$2x-2r\cos\theta+2yy'-2r\sin\theta y'=0$$
$$1+yy''+(y')^2-r\sin\theta y''=0$$
$$\tag3r\sin\theta=\frac{1+yy''+(y')^2}{y''}$$
Substitute $(3)$ in $(1)$,
$$(x-r\cos\theta)^2+(y-r\sin\theta)^2=r^2$$
$$x^2-2rx\cos\theta+y^2-2ry\sin\theta+r^2\sin^2\theta=r^2\sin^2\theta$$
$$x^2-2rx\cos\theta+y^2-2ry\sin\theta=0$$
$$\tag4r\cos\theta=\frac{x^2+y^2-2ry\sin\theta}{2x}$$
Now substitute $(3)$ in $(4)$ and then $(3)$ and $(4)$ in $(2)$.
It is likely that there may be an easier way than this.
A: For a circle polar coordinates is a natural choice. It avoids differential relations between $ (x,\theta) $. It is convenient to find a pure polar equation of a circle of arbitrary radius (or diameter D ) whose inclination $ \alpha$ to a reference direction like x-axis is arbitrary. Curvature is a second order differential and hence these arbitrary constants $ ( D, \alpha) $ need to be involved and later on eliminated in differentiation:
From the sketch it is seen they are relatively positioned and geometrically related as:
$$    r = D  \cos(\theta + \alpha) $$ 
whose differential equation is the well known:
$$ \boxed {\frac{d^2 r}{d \theta^{2}} + r =0.} $$
Rectangular coordinates to be computed after numerical integration.
EDIT 1:
In rectangular coordinates we can derive as follows.  $\alpha$ can be taken wlog as zero as seen above, anyhow it disappears.
$$ x = D \cos ^2\theta=\frac{D\,x^2}{x^2+y^2} $$
$$ x^2 + y^2 = D\,x $$
Differentiating once
$$ x + y\,y^{\prime} = D/2 $$
Differentiate again
$$\boxed {y\, y^{\prime \,\prime} + 1 + y^{\prime\, 2 }=0.}$$
One disadvantage with rectangular coordinates numerical integration is that it cannot proceed beyond point of the vertical tangent.. unless other tricks are used.

A: The slope of the line segment joining the origin to a point $(x,y)$ on the circle is $y/x$. The tangent at the point $(x,y)$ is orthogonal to this line segment and hence has slope $-x/y$. So you get the equation $y'=-x/y$ whose integral curves are circles are concentric circles centered about the origin.
