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. . .