We all know that $x^2+y^2=r^2$ is a circle. What does $(x^2+y^2)^2$ signify? In general, what is $(x^2+y^2)^n$?

  • $\begingroup$ It remains a circle... assuming that $x,y\in\mathbb{R}$. The reason behind it is simply given by the fact that you can always take the n-th root on both sides and you don't change the possibile values... $\endgroup$ – b00n heT Jan 25 '14 at 14:08
  • $\begingroup$ This doesn't answer your question, but it's interesting to note that if $x^2 + y^2 - r^2 = 0$ is squared on both sides, then the resulting polynomial still determines a circle, but all the points are multiplicity two roots of the equation now. $\endgroup$ – Nick Jan 25 '14 at 14:09
  • $\begingroup$ Perhaps you're thinking of the "superellipses", a.k.a. "(even-degree) Fermat curves" $x^{2n} + y^{2n} = 1$, which become "successively more square" as $n \to \infty$...? $\endgroup$ – Andrew D. Hwang Jan 25 '14 at 14:56
  • $\begingroup$ @ hAcKnRoCk : Sorry I cannot understand your question because on one hand you write x²+y²=r² which is an equation and on the other hand (x²+y²)² which is not an equation (there is no = in it). What exactly do you want to compare ? $\endgroup$ – JJacquelin Jan 25 '14 at 15:12
  • 1
    $\begingroup$ Higher dimensions = more variables. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 25 '14 at 15:59

Using the transformation: $$x=r\cos(t),~~y=r\sin(t)$$ there is no change. I mean the circle remains unchanged.

  • $\begingroup$ Thanks. But i am asking only about the LHS. I have edited the question to make it more clear $\endgroup$ – hAcKnRoCk Jan 26 '14 at 10:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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