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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.

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

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    @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).2012-01-22

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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

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

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    True, but I think showing the steps in getting there is at least as important as the result.2012-01-22
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This curve is the Lemniscate of Bernoulli.

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

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