Q.Find the differential equation of all the circles that pass through origin.

A. $\displaystyle (x^2 + y^2)y'' = 2(xy' - y)(1 + (y')^2) $

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- Mar 9th 2008, 03:02 AMAltairSolution required
Q.Find the differential equation of all the circles that pass through origin.

A. $\displaystyle (x^2 + y^2)y'' = 2(xy' - y)(1 + (y')^2) $ - Mar 9th 2008, 03:36 AMmr fantastic
Well, I have a plan but you'll need to put in the details.

If the circle $\displaystyle (x - h)^2 + (y - k)^2 = r^2$ passes through the origin then $\displaystyle h^2 + k^2 = r^2$.

Then the equation of the circle becomes

$\displaystyle x^2 - 2xh + y^2 - 2yk = 0$ .... (1)

Differentiate (1) with respect to x:

$\displaystyle 2x - 2h + 2y \, y' - 2k \, y' = 0$ .... (2)

Differentiate (2) with respect to x:

$\displaystyle 2 + 2 (y')^2 + 2 y \, y'' - 2k \, y'' = 0$ .... (3)

Clearly k needs to be eliminated from (3).

(1) and (2) can be solved simultaneously to get k and h in terms of x, y and y'. You want k. Substitute the result into (3). - Mar 10th 2008, 09:49 AMAltair
Need help with substitution. It gets too complicated at the end.

- Mar 10th 2008, 11:58 AMmr fantastic