Find a polar representation for a curve. I have the following curve:
$(x^2 + y^2)^2 - 4x(x^2 + y^2) = 4y^2$ and I have to find its polar representation.
I don't know how. I'd like to get help .. thanks in advance.
 A: Just as the Cartesian has two variables, we will have two variables in polar form:
$$x = r\cos \theta,\;\;y = r \sin \theta$$
We can also use the fact that $x^2 + y^2 = (r\cos \theta)^2 + (r\sin\theta)^2 = r^2 \cos^2\theta + r^2\sin^2 \theta = r^2\underbrace{(\sin^2 \theta + \cos^2 \theta)}_{= 1} =r^2$
This gives us $$r^4 - 4r^3\cos \theta - 4r^2 \sin^2\theta = 0 \iff r^2 - 4r\cos\theta - 4\sin^2\theta = 0$$
A: Replace $x$ by $r\cos\theta$, $y$ by $r\sin\theta$, and simplify. Simplification is in principle not obligatory, but it would be a little strange, for example, not to replace $r^2\cos^2\theta+r^2\sin^2\theta$ by $r^2$.
In some cases, one can express the equation we get in the form $r=f(\theta)$, where $f$ is given by an explicit formula. If it can be done in a fairly simple way, it is generally considered desirable to do so.
Added: Let us do this with your example. Do the substitution mechanically, noting that $x^2+y^2=r^2$. We get $r^4-4r^3\cos\theta=4r^2\sin^2\theta$. This is correct, but we might want to do further processing. Divide through by $r^2$, remembering that this throws away the possibility $r=0$, that is, the origin. 
Now replace $\sin^2\theta$ by $1-\cos^2\theta$. We get $r^2-4r\cos\theta=4-4\cos^2\theta$. We could now solve the quadratic either for $r$ or for $\cos\theta$. By completing the square, we get $(r-2\cos\theta)^2=4$. By the way, note that $\theta=0$, $r=0$ satisfies this equation, so we have not thrown away the origin, and need not feel guilty about dividing by $r^2$.
This is simple enough, but if we wish, note that $r-2\cos\theta=\pm 2$, so $r=2\cos\theta\pm 2$. However, for many presentations (but not all) of polar coordinates, we cannot have negative $r$, so $r=2\cos\theta+2$.
Remark: Your particular problem simplified dramatically. But that is because the numbers were carefully chosen. (We could also have started by simplifying the original rectangular coordinates equation.)
