singly
(0, -b 2) and (under the radical sign -a/2(+-) (b 2+a 2/4))
The center of this circle must be
(-a/2,y)
According to the equal distance between the center of the circle and these three points,
Get the position of the center of the circle as follows
(-a/2,( 1-b^2)/2)
Then list the equations of the circle,
Simplify to obtain
x^2+ax+y^2+b^2y-y-b^2=0
Because a and b can take arbitrary values.
Therefore, x can be 0,-a.
Y can be 1.
Therefore, the fixed point is (0, 1).
It died,
No, please forgive me. . .