If one of the diameters of the circle ${x^2} + {y^2} - 2\sqrt 2 x - 6\sqrt 2 y + 14 = 0$ is a chord of the circle ${(x - 2\sqrt 2 )^2} + {(y - 2\sqrt 2 )^2} = {r^2}$, then the value of r² is equal to ____________.