jmc

Rearranging $y=2x+c$ gives $2x=y-c.$ Substituting into $y^2=8x,$ we get $$y^2=4(y-c)=4y-4c,$$or $y^2-4y+4c=0.$ Since we have a tangent, this quadratic will have a double root. In other words, its discriminant will be 0. Hence, $(-4{)}^2-4(4c)=16-16c=0,$ which means $c=\boxed{1}.$