Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

share|improve this question
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
1  
@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
show 1 more comment

3 Answers

up vote 3 down vote accepted

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

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

share|improve this answer
    
+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
add comment

This curve is the Lemniscate of Bernoulli.

share|improve this answer
add comment

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

enter image description here

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.