imc

Since points on the graph make the equation true, substitute $\sqrt{\pi }$ in to the equation and then solve to find $a$ and $b$. $y^2+{\sqrt{\pi }}^4=2{\sqrt{\pi }}^{2}y+1$ $y^2+{\pi }^2=2\pi y+1$ $y^2-2\pi y+{\pi }^2=1$ $(y-\pi {)}^2=1$ $y-\pi =\pm 1$ $y=\pi +1$ $y=\pi -1$ 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. $|(\pi +1)-(\pi -1)|=2$ The answer is $\boxed{2}$