# How to normalize this circle equation?

I am given a circle described by the equation below. Is there any way I can bring it to the form $(x-a)^2 + (y-b)^2 = c^2$ to have it be normal? My intent is to translate it to polar coordinates and I think I'd get much nicer results if I could normalize the equation.

$$(x^2 + y^2)^2 = 9(x^2 - y^2)$$

-
That's no circle. It has fourth powers in it (left-hand side). – msh210 Jan 22 '12 at 19:20
To persuade yourself that it is not a circle, ask a program (say Wolfram Alpha) to graph it. – André Nicolas Jan 22 '12 at 19:23
The translation to polar coordinates simply gives you $r^2=9\cos 2\theta$. – Brian M. Scott Jan 22 '12 at 19:26
@msh210 The exercise actually says that it describes a cylinder; was I wrong in thinking that if that were true, it would also describe a circle? (As for the equation, I copied it correctly, I guess whoever copied it before me might have made a mistake.) – Paul Manta Jan 22 '12 at 19:27
@PaulManta: "Cylinder" is sometimes used to describe any solid with congruent cross-sections perpendicular to an axis (with the more specific "right circular cylinder" for the kind with congruent circular cross-sections whose centers all lie on a line perpendicular to the planes of the cross-sections). – Isaac Jan 22 '12 at 19:30

As I stated in a comment, that's not the equation of a circle. But if your intent is to rewrite it in polar coordinates, just substitute by $x=r\cos\theta,y=r\sin\theta$: then $(x^2+y^2)^2=9(x^2-y^2)$ becomes

$$\begin{array}{rcl}(r^2\cos^2\theta+r^2\sin^2\theta)^2&=&9(r^2\cos^2\theta-r^2\sin^2\theta)\\(r^2)^2&=&9(r^2\cos(2\theta)\\r^2&=&9\cos(2\theta)\end{array}$$

-
+1. I was just about to edit this into my answer when you posted it. – Isaac Jan 22 '12 at 19:27
And I see now that Brian M. Scott beat me to this solution by half a minute or so (in a comment on the question). – msh210 Jan 22 '12 at 19:30
True, but I think showing the steps in getting there is at least as important as the result. – Isaac Jan 22 '12 at 19:31

This curve is the Lemniscate of Bernoulli.

-

As msh210 said in a comment, it's not a circle.

-