The shortest distance between the curves $x^2+y^2=8$ and $2[{(x-3)}^2+{(y+3)}^2]={(x-y-2)}^2$ is $k$ unit, then $k$ is equal to

Second curve is parabola with focus $(3,-3)$ and directrix $x-y-2=0$ equation of axis is $x+y=0$ and foot of directrix $(1,-1)$ Vertex = $(2,-2)$
$\therefore$ Shortest distance = 0