What is the graph of $y^4+1=x^4+2y^2$ in the coordinate plane?
$\textbf{(A) } \text{two intersecting parabolas} \qquad \textbf{(B) } \text{two nonintersecting parabolas} \qquad \textbf{(C) } \text{two intersecting circles} \qquad \\\\ \textbf{(D) } \text{a circle and a hyperbola} \qquad \textbf{(E) } \text{a circle and two parabolas}$
Since the equation has even powers of $x$ and $y$, let $y'=y^2$ and $x' = x^2$. Then $y'^2 + 1 = x'^2 + 2y'$. Rearranging gives $y'^2 - 2y' + 1 = x'^2$, or $(y'-1)^2=x'^2$. There are two cases: $y' \leq 1$ or $y' > 1$.
If $y' \leq 1$, taking the square root of both sides gives $1 - y' = x'$, and rearranging gives $x' + y' = 1$. Substituting back in $x'=x^2$ and $y'=y^2$ gives us $x^2+y^2=1$, the equation for a circle.
Similarly, if $y' > 1$, we take the square root of both sides to get $y' - 1 = x'$, or $y' - x' = 1$, which is equivalent to $y^2 - x^2 = 1$, a hyperbola.
Hence, our answer is $\boxed{\textbf{(D)}\ \text{a circle and a hyperbola}}$.
~[Bxiao31415](https://artofproblemsolving.com/wiki/index.php/User:Bxiao31415)