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(1) Since the left sides are identical in given equations of the circles, their right sides are equal for intersection points
x = y.
(2) Substituting x=y into the circles equations gives
x^2 + x^2 = 2x ====> 2x^2 = 2x ====> 2x^2 - 2x = 0 ====> 2x(x-1) = 0.
The roots are x= 1 and x= 0, so the intersection points are A=(1,1) and B=(0,0).
(3) The center of the third circle lies on the perpendicular bisector to the segment AB.
An equation of this perpendicular bisector is y-0.5 = -(x-0.5) = -x + 0.5.
(4) At the same time, the center of the third circle lies on the line y= 2.
So, the center of the third circle is at
2 - 0.5 = -x + 0.5,
which implies x= -1.
Thus the center of the third circle is the point P = (-1,2).
The radius of the third circle is the distance of the point P from the origin (0,0), i.e. = = .
(5) Thus the equation of the circle under the question is
+ = ,
or, equivalently,
+ = 5. ANSWER