# Polar equation to cartesian

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

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

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

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

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.

thenewbosco
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?

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