How can I convert a polar equation into a cartesian equation with two circles?

[thenewbosco]

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

 

[Tide]

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

 

[benorin]

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]

Well, i get something like

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

from which the conclusion follows easily.

Daniel.

 

[thenewbosco]

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]

For starters, rewrite it in the form:
$$(x^{2}+y^{2})^{2}=(x+y)^{2}$$

 

[arildno]

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]

There , or rather not there: all better?