Sketching the graph of a fourth degree polynomial curve

Question

My approach doesn't seem to work.

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

Any suggestions and solutions would be appreciated .

Answer 1

Writing $u=x^2$ and $v=y^2$ we get the equation $$(v-2)^2+(u-2)^2 = 2,$$ so we know that $(x^2,y^2)$ must lie on the upper-right quarter of the circle of radius $\sqrt{2}$ centered in $(2,2)$. To get the graph (or a good approximation thereof), you can draw this quarter of circle, and then take the square root(s) of the coordinates of its points to get points of your graph.