设AB:y=kx+m,A(x1,y1)B(x2,y2)则有：y1=kx1+m,y2=kx2+m由于AB于圆相切则圆心(0,0)到AB距离为半径即：√(2/3)=|k*0-0+m|/[√(1+k^2)]得：2k^2+2=3m^2 ----(1)联立y=kx+m与x^2+2y^2=2得：(1+2k^2)x^2+4kmx+2m^2-2=0则...