Question

Help!!!
Find the differential equation of all the circles with centers x + 2y = 0 and passing through (0,0).

Answer 1

How about this...
A circle with center (-2u, u), radius r is written as
\[(x + 2u)^{2} + (y - u)^{2} = r^{2}\]
Differentiating both side with respect to x, we get
\[ x + 2u + (y - u)y' = 0\]

Answer 2

how do you get the center?

Answer 3

This circle
\[ (x+2u)^2+(y−u^)2=r^2\]
has its centre at (-2u, u), right?

Answer 4

the centers of the all circles are lying at a line x + 2y = 0. how do you get the center (-2u,u) and the circle must passing through (0,0).

Answer 5

Let x = -2u and y = u. For any u, (-2u) + 2u = 0. So (-2u, u) is on the line x + 2y = 0.
I got this pair (-2u, u) just by modifying x + 2y = 0 -> x = -2y.

Answer 6

ok, thanks

Answer 7

I'm glad if it helped. Thanks.