(1)圆C:x2+(y-1)2=5的圆心为C(0,1),半径为. ∴圆心C到直线l:mx-y+1-m=0的距离d=≤=< ∴直线l与圆C相交; (2)由直线方程mx-y+1-m=0,得m(x-1)-y+1=0,可知直线l过定点P. 当M与P不重合时,连结CM、CP,则CM⊥MP, ∴|CM|2+|MP|2=|CP|2 设M(x,y)(x≠1),则x2+(y-1)2+(x-1)2+(y-1)2=1, 化简得:x2+y2-x-2y+1=0(x≠1); 当M与P重合时,x=1,y=1也满足上式. 故弦AB中点的轨迹方程是x2+y2-x-2y+1=0. (3)设A(x1,y1),B(x2,y2),由=,得=, ∴1-x1=(x2-1),化简的x2=3-2x1…① 又由,消去y得:(1+m2)x2-2m2x+m2-5=0…(*) ∴x1+x2=…② 由①②解得x1=,代入(*)式解得m=±1, ∴直线l的方程为x-y=0或x+y-2=0. |