+y2)y=0.Р故线段AB是圆C的直径.Р证法二:∵||=||,Р∴()2=()2,Р即··,Р整理得·=0.Р∴x1x2+y1y2=0. ①Р若点(x,y)在以线段AB为直径的圆上,则Р·=-1(x≠x1,x≠x2).Р去分母得(x-x1)(x-x2)+(y-y1)(y-y2)=0.Р点(x1,y1),(x1,y2),(x2,y1),(x2,y2)满足上方程,展开并将①代入得Рx2+y2-(x1+x2)x-(y1+y2)y=0.Р∴线段AB是圆C的直径.Р证法三:∵||=||,Р∴()2=()2,即Р··,Р整理得·=0.Р∴x1x2+y1y2=0. ①Р以AB为直径的圆的方程是Р(x-)2+(y-)2=[(x1-x2)2+(y1-y2)2].Р展开,并将①代入得Рx2+y2-(x1+x2)x-(y1+y2)y=0.Р∴线段AB是圆C的直径.Р(2)解法一:设圆C的圆心为C(x,y),Р则Р∵=2px1,=2px2(p>0).Р∴x1x2=.Р又∵x1x2+y1y2=0,Р∴x1x2=-y1y2.Р∴-y1y2=.Р∵x1x2≠0,∴y1y2≠0.Р∴y1y2=-4p2.Р∴x==()=(+2y1y2)-Р=(y2+2p2).Р∴圆心的轨迹方程为y2=px-2p2.Р设圆心C到直线x-2y=0的距离为d,则Рd=Р=Р=.Р当y=p时,d有最小值,Р由题设得=.Р∴p=2.Р解法二:设圆C的圆心为C(x,y),Р则.Р∵=2px1,=2px2(p>0),Р∴x1x2=.Р又∵x1x2+y1y2=0,Р∴x1x2=-y1y2.Р∵x1x2≠0,∴y1y2=-4p2.Р∵x==()Р=(+2y1y2)-Р=(y2+2p2).Р∴圆心的轨迹方程为y2=px-2p2.Р设直线x-2y+m=0与x-2y=0的距离为,Р则m=±2.Р∵x-2y+2=0与y2=px-2p2无公共点,
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