zxqsyr2010-01-28TA获得超过14.3万个赞关注设A(x1,y1),B(x2,y2),两点在圆上:x1^2+y1^2=1x2^2+y2^2=1k=(y1-y2)/(x1-x2)相减整理得:(x1+x2)+(y1+y2)k=0设C为AB中点(x,y),由上得:x+ky=0kx-y+1=0联立消去k,可解得其轨迹为:x^2+(y-1/2)^2=1/4