设圆心坐标M(0,m),圆的方程为：x^2+(y-m)^2=1
（1）求直线与圆的交点：x^2 + ( 2x+2-m)^2=1,化简为：
5x^2 + 4(2-m)x + (3-m)(1-m) = 0
x1+x2 = 4(m-2)/5；x1*x2 = (3-m)(1-m)/5；
所以 (x1-x2)^2 = (x1+x2)^2 - 4x1*x2 = 16(m-2)^2/25 - 4(3-m)(1-m)/5 = -4(m^2-4m-1)/25
同理：(y1-y2)^2 = -16(m^2-4m-1)/25
弦长：(x1-x2)^2 + (y1-y2)^2 = 16/5 = -20(m^2-4m-1)/25
解得：m=1 或3.因为圆心在直线下方：m