Polar equation to cartesian

i have this equation: $$r=\sqrt{1+sin2\theta}$$
and am to convert to cartesian equation and from the equation see that it consists of two circles and directly note the radii of the cirlcles from the equation.

so far i have manipulated it and gotten:

$$x^6 + 3x^4 y^2 + 3 x^2 y^4 + y^6 - x^4 -6x^2 y^2 - 4x^3 y -y^4 - 4y^3 x=0$$

any suggestions on how to get it into the required form?
thanks

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Tide
Homework Helper
HINT: $$\sin \alpha = \cos \left (\frac {\pi}{2} - \alpha\right)$$

benorin
Homework Helper
It should be a fourth order equation in x and y. I get:

$$\left[ \left(x+\frac{1}{2}\right) ^2 + \left(y+\frac{1}{2}\right) ^2 -\frac{1}{2}\right] \left[ \left(x-\frac{1}{2}\right) ^2 + \left(y-\frac{1}{2}\right) ^2 -\frac{1}{2}\right] =0$$

dextercioby
Homework Helper
Well, i get something like

$$x^2 +y^2 =1+\frac{2xy}{x^2 +y^2}$$,

from which the conclusion follows easily.

Daniel.

dextercioby said:
Well, i get something like
$$x^2 +y^2 =1+\frac{2xy}{x^2 +y^2}$$,
from which the conclusion follows easily.
Daniel.
yes, i too found this but can you explain how the conclusion follows easily from this?

arildno
Homework Helper
Gold Member
Dearly Missed
For starters, rewrite it in the form:
$$(x^{2}+y^{2})^{2}=(x+y)^{2}$$

arildno
Homework Helper
Gold Member
Dearly Missed
benorin said:
then factor it using difference of squares
That was to be my next hint..

Then, the grand finale would have been to refer to your earlier solution to this problem. I won't do that now.

benorin
Homework Helper
There , or rather not there: all better?