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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 $$r^2$$ is equal to
Correct Answer: 10
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