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

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

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.
