设圆心坐标M(0,m),圆的方程为：x^2+(y-m)^2=1\x0d（1）求直线与圆的交点：x^2 + ( 2x+2-m)^2=1,化简为：\x0d5x^2 + 4(2-m)x + (3-m)(1-m) = 0\x0dx1+x2 = 4(m-2)/5；x1*x2 = (3-m)(1-m)/5；\x0d所以(x1-x2)^2 = (x1+...