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My approach doesn't seem to work.

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

Any suggestions and solutions would be appreciated .

hardmath
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Abu Bardewa
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    And what is your approach...? – CiaPan Dec 08 '15 at 11:52
  • But these things apply to a function. And what's more, to a function of one variable. You don't have an explicit function here, that is an equation with two variables. We might consider it an implicit function, but I think you need to state that (and try to extract an explicit function, if possible, or at least decide if it is $y$ as a function of $x$ or $x$ as a function of $y$). – CiaPan Dec 08 '15 at 11:58
  • Try to find a domain first. What values of $x$ and what values of $y$ may, and what values can't satisfy the equation? – CiaPan Dec 08 '15 at 11:59
  • YEh, I noticed that . I thought expanding brackets would be too tiresome . So. – Abu Bardewa Dec 08 '15 at 12:00
  • If you forget the squares for $x, y$, you get a circle around $(2,2)$ with radius $\sqrt 2$. All those points have to be replaced by their square root, and can be positive or negative (so we have 3 mirror images). And otherwise, just wolfram it: http://www.wolframalpha.com/input/?i=%28y%5E2%E2%88%922%29%5E2%2B%28x%5E2%E2%88%922%29%5E2%3D2 – Pieter21 Dec 08 '15 at 12:21

1 Answers1

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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.

Daniel Robert-Nicoud
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