Since points on the graph make the equation true, substitute $\sqrt{\pi}$ in to the equation and then solve to find $a$ and $b$.  $$ \begin{aligned} \&y^{2}+\sqrt{\pi}^{4}=2 \sqrt{\pi}^{2} y+1 \textbackslash\textbackslash{} \&y^{2}+\pi^{2}=2 \pi y+1 \textbackslash\textbackslash{} \&y^{2}-2 \pi y+\pi^{2}=1 \textbackslash\textbackslash{} \&(y-\pi)^{2}=1 \textbackslash\textbackslash{} \&y-\pi=\pm 1 \textbackslash\textbackslash{} \&y=\pi+1 \textbackslash\textbackslash{} \&y=\pi-1 \end{aligned} $$  There are only two solutions to the equation $(y-\pi)^{2}=1$, so one of them is the value of $a$ and the other is $b$. The order does not matter because of the absolute value sign. $$ \textbar(\pi+1)-(\pi-1)\textbar=2 $$ 
Points $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ are distinct points on the graph of $y^{2}+x^{4}=2 x^{2} y+1$. What is $\textbar a-b\textbar$? 
(1) $$1$$ 
(2) $\frac{\pi}{2}$ 
(3) $$2$$ 
(4) $\sqrt{1+\pi}$ 
(5) $1+\sqrt{\pi}$ 