smc

Since the equation has even powers of $x$ and $y$, let $y^{\prime }=y^2$ and $x^{\prime }=x^2$. Then $y^{\prime 2}+1=x^{\prime 2}+2y^{\prime }$. Rearranging gives $y^{\prime 2}-2y^{\prime }+1=x^{\prime 2}$, or $(y^{\prime }-1{)}^2=x^{\prime 2}$. There are two cases: $y^{\prime }\le 1$ or $y^{\prime }>1$. If $y^{\prime }\le 1$, taking the square root of both sides gives $1-y^{\prime }=x^{\prime }$, and rearranging gives $x^{\prime }+y^{\prime }=1$. Substituting back in $x^{\prime }=x^2$ and $y^{\prime }=y^2$ gives us $x^2+y^2=1$, the equation for a circle. Similarly, if $y^{\prime }>1$, we take the square root of both sides to get $y^{\prime }-1=x^{\prime }$, or $y^{\prime }-x^{\prime }=1$, which is equivalent to $y^2-x^2=1$, a hyperbola. Hence, our answer is $\boxed{\text{a circle and a hyperbola}}$.