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I was recently asked to sketch $(x^2-2)^2+(y^2-2)^2=2$, which did not prove to be too problematic, for establishing the range and domain of the expression gives nearly all of it away.

I then asked myself what would happen as I change the constant on the R.H.S. I recommend you try this yourself:

https://www.desmos.com/calculator/tlksmrpzxu

The constant that stood out as yielding the most intriguing result is $4$. The graph looks like this:

$(x^2-2)^2+(y^2-2)^2=4$

I then wondered how on earth I could deduce this just by analysing $(x^2-2)^2+(y^2-2)^2=4$.

One of my approaches involved spotting that the two 'orbits' are in the shape of ellipses with equations (derived partly experimentally, partly using the ellipse formula) $x^2-xy\sqrt{2}+y^2=2$ and $x^2+xy\sqrt{2}+y^2=2$. Can I derive these just from $(x^2-2)^2+(y^2-2)^2=4$?

Plato
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    Interesting. $(x^2-a)^2+(y^2-a)^2-a^2 = (x^2+y^2-a+xy\sqrt 2) (x^2+y^2-a-xy\sqrt 2)$. I have no clue how one would go from left hand side to the right without knowing the factorization holds beforehand. – Ennar Nov 23 '17 at 16:53

2 Answers2

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Well, you can go backwards, and multiply $x^2 - \sqrt2xy+y^2-2$ and $x^2 + \sqrt2xy+y^2-2$ to get the product $x^4-4x^2+y^4-4y^2 + 4$, equal to $(x^4-4x^2+4) + (y^4-4y^2+4)-4$, but I did not see the reverse path when I looked at your question.

Lubin
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More generally,

$(x^2-a)^2+(y^2-a)^2-b = (x^2-Axy+y^2-B)(x^2+Axy+y^2-B)$ iff $A^2=2, B=a, b=a^2$

but like @Lubin I can't see this at once.

lhf
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