Two circles $${x^2} + {y^2} = 6$$ and $${x^2} + {y^2} - 6x + 8 = 0$$ are given. Then the equation of the circle through their points of intersection and the point (1, 1) is

A square is inscribed in the circle $${x^2} + {y^2} - 2x + 4y + 3 = 0$$. Its sides are parallel to the coordinate axes. The one vertex of the square is