Then x=t-y

Substitute into the original equation

→y^2-2ty+t^2+y^2-2y+ 1-4=0

2y^2-(2t+2)y+t^2-3=0

This equation should have a solution.

Then the discriminant △/4 = t 2+2t+ 1-2t 2+6.

=-t^2+2t+7≥0

→t^2-2t-7≤0

The maximum value is 2√2 and the minimum value is 1-2√2.