已知圆M的圆心M在Y轴上,半径为1.直线L：y=2x+2被圆M所截得弦长为（4√5）/5,且圆心M在直线L的下方

设圆心坐标M(0,m),圆的方程为：x^2+(y-m)^2=1x0d（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) = 0x0dx1+x2 = 4(m-2)/5；x1*x2 = (3-m)(1-m)/5；x0d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)/25x0d同理：(y1-y2)^2 = -16(m^2-4m-1)/25x0d弦长：(x1-x2)^2 + (y1-y2)^2 = 16/5 = -20(m^2-4m-1)/25x0d解得：m=1 或3.因为圆心在直线下方：m