1．从圆外一点P(1,1)向圆x2+y2=1引割线,交该圆于A,B两点,求弦AB的中点的轨迹方程.

设斜率是ky-1=k(x-1)y=kx+(1-k)代入(k²+1)x²+2k(1-k)x+(1-k)²-1=0x1+x2=-2k(1-k)/(k²+1)y=kx+(1-k)所以y1+y2=k(x1+x2)-2(1-k)=2(1-k)/(k²+1)中点x=(x1+x2)/2,y=(y1+y2)/2所以x/y=-k而y-1=k(x-1)所以k=(y-1)/(x-1)所以x/y=-(y-1)/(x-1)-y²+y=x²-xx²+y²-x-y=0(x-1/2)²+(y-1/2)²=1/2直线是割线圆心(0,0)到直线距离小于半径直线kx-y+(1-k)=0所以|1-k|/√(k²+1)