圆方程可变为 (x-1)^2 + (y+3)^2 = 10 -5a图心(1,-3)该圆关于直线y=x+2b成轴对称图形,图心必然在该直线上:-3 = 1 + 2bb = -2圆存在(x-1)^2 + (y+3)^2 > 0,10-5a > 0,a < 2a-b < 2 - b = 2-(-2) = 4,a < 4...