If one 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 \( r^{2} \) is equal to \( \_ \_ \_ \_ \_ \_ \).
