Find the equation of the circle that passes through the point of intersection of the circles x^2+y^2=2x, x^2+y^2=2y and contains the point (2,1)

Answer 1

change your equations to
(x-1)^2+y^2=1
x^2+(y-1)^1=1

The radius is one, and both circles are offset from the origin by 1, so the common point is at the origin (0,0). So now, what circle passes thru the origin and 2,1? Answer; many, it takes three points to determine the circle. If the question meant "passes thru the points of intersection"...
O,0 is one point
the other...
subtracting the second equation from the first.
0=2x-2y or x=y
2y^2=2y again, 0 is one solution, the other is y=1, in which case x^2+1=2 x^2=1 or x=+-1, and the first equation is not comparable with x=-1
second intersection then (1,1), (2,1), (0,0) Now three points, this defines a circle. The easy way to find the center of the circle is to find a point (a,b) which is equidistant from all these points.
(x-h)^2+(y-k)^2=r^2