jmc

Completing the square in $x$ and $y,$ we get $${\left(x-\frac{1}{2}\right)}^2+{\left(y-\frac{1}{2}\right)}^2=\frac{1}{2}.$$This represents the equation of the circle with center $\left(\frac{1}{2},\frac{1}{2}\right)$ and radius $\frac{1}{\sqrt{2}}.$ Hence, the largest possible value of $x$ is $\frac{1}{2}+\frac{1}{\sqrt{2}}=\boxed{\frac{1+\sqrt{2}}{2}}.$