Because the center of the circle C must be on the perpendicular line of AB, finding the intersection point between the perpendicular line of AB and the straight line y=x is the demand.

The slope of line AB is: kAB=(3-2)/(0-3)=- 1/3.

Therefore, the slope k of the perpendicular line of AB =-1/(-1/3) = 3.

The midpoint of AB is (3/2, 5/2)

So the vertical line of AB is: y-5/2=k(x-3/2)=3(x-3/2)=3x-9/2, that is, y=3x-2.

The intersection with the straight line y=x is C( 1, 1).

r=ac=√[(0- 1)^2+(3- 1)^2]=√5

So the equation of circle C is: (X- 1) 2+(Y- 1) 2 = 5.