直线过点（-1,1）∴可以设直线方程为：y-1=k(x+1),即y=kx+k+1,代入圆的方程X^2+y^2-2x=0得：x^2+(kx+k+1)^2-2x=0x^2+k^2x^2+2k(k+1)x+(k+1)^2-2x=0(k^2+1)x^2+2(k^2+k-1)x+(k+1)^2=0x1+x2=-2(k^2+k-1)/(k^2+1)x1x2=(...