已知直线x+根号3y-m=0与圆x2+y2=1交于A,B两点,则向量OA+向量OB共线的向量为?

x^2+y^2=1x=m-√3y,即：(m-√3y)^2+y^2=1即：4y^2-2√3my+m^2-1=0即：y1=(√3m+sqrt(4-m^2))/4,y2=(√3m-sqrt(4-m^2))/4故：x1=m-√3y1,x2=m-√3y2即AB中点坐标：x0=(x1+x2)/2=m-√3(y1+y2)/2=m-3m/4=m/4y0=(y1+y2)/...