<s>[INST] 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?
 [/INST] 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.
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