The trajectory equation of the center of moving circle X05 + Y05 - (4m + 2) - 2My + 4m05 + 4m + 1 = 0 is

The trajectory equation of the center of moving circle X05 + Y05 - (4m + 2) - 2My + 4m05 + 4m + 1 = 0 is

Is 05 quadratic?
The coordinates of the center of the circle are x = 2m + 1, y = M,
The elimination of M leads to x = 2Y + 1,
That is x-2y-1 = 0. This is the trajectory equation of the center

If the center of the circle x2 + Y2 - (4m + 2) x-2my + 4m2 + 4m + 1 = 0 is on the straight line x + y-4 = 0, then the area of the circle is

x2+y2-(4m+2)x-2my+4m2+4m+1=0
It can be reduced to (x-2m-1) ^ 2 + (y-m) ^ 2 = m ^ 2,
The center of the circle (2m + 1, m) is on X + y-4 = 0,
Then 2m + 1 + M-4 = 0, M = 1,
The radius of the circle is m = 1 and the area of the circle is π