Question 1143657
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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.  {{{sqrt((-1)^2+2^2)}}} = {{{sqrt(1+4)}}} = {{{sqrt(5)}}}.


(5)  Thus the equation of the circle under the question is


         {{{(x-(-1))^2}}} + {{{(y-2)^2}}} = {{{(sqrt(5))^2}}},

     or, equivalently,


         {{{(x+1)^2}}} + {{{(y-2)^2}}} = 5.    <U>ANSWER</U>
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