设直线方程为y=k(x-2),代入圆方程得x^2+k^2*(x-2)^2=1,也即(1+k^2)x^2-4k^2*x+4k^2-1=0 ①设交圆于两点A(x1,y1),B(x2,y2),有x1+x2=4k^2/(1+k^2)y1+y2=k(x1-2)+k(2x-2)=k(x1+x2)-4k=4k^3/(1+k^2)-4k=-4k/(1+k^2)于是...