How to sketch $(y^2-2)^2+(x^2-2)^2=2$?

Question

I get the overall idea, but why is the shape of these figures not a circle?

$(x-2)^2+(y-2)^2=2$ is a circle. A quartic equation normally does not have a circular graph. — Zhuoran He, Nov 05 '17 at 19:41
this Looks nice https://www.wolframalpha.com/input/?i=plot+(y%5E2-2)%5E2%2B(x%5E2-2)%5E2%3D2 — Dr. Sonnhard Graubner, Nov 05 '17 at 19:43
@DonaldSplutterwit, thanks for your link. I got a pillow shape using $(x^2-2)^2+(y^3-2)^2=7$. — Zhuoran He, Nov 05 '17 at 19:45

score 3 · Answer 1 · edited Nov 05 '17 at 19:53

Consider new frame with coordinate axes defined as $X=x^2$ and $Y=y^2$.

In this new frame of reference, the equation you mentioned will be a perfect circle with center at (2,2) and radius of $\sqrt 2$.

But every point on this new frame of reference will have, 4 copies in the original $x,y$ frame. Why?
A point $(X_0,Y_0)$ in new frame will have $4$ copies which will be $(\sqrt {X_0},\sqrt {Y_0}),(\sqrt {X_0},-\sqrt {Y_0}),(-\sqrt {X_0},\sqrt {Y_0}),(-\sqrt {X_0},-\sqrt {Y_0}) $.

Since the circle in new reference frame lies completely in 1st quadrant, therefore the circle will be replicated into $4$ figures in $4$ different quadrants in original frame of reference.

This can also be verified by viewing the graph here.

But why are these "circles" pillow-shaped and not normal ones? — John Robbers, Nov 08 '17 at 21:21

How to sketch $(y^2-2)^2+(x^2-2)^2=2$?

1 Answers1