jmc

We can write $$\begin{array}{rl}2x^2+2xy+y^2-2x+2y+4& =(x^2+y^2+1+2x+2y+2xy)+(x^2-4x+4)-1\\ & =(x+y+1{)}^2+(x-2{)}^2-1.\end{array}$$Thus, the minimum value is $\boxed{-1},$ which occurs when $x+y+1=0$ and $x-2=0,$ or $x=2$ and $y=-3.$