过Q(2,-4)作圆O:x^2+y^2=9的割线,交圆O于点A,B.求AB中点P的轨迹方程
问题描述：
过Q(2,-4)作圆O:x^2+y^2=9的割线,交圆O于点A,B.求AB中点P的轨迹方程
RT
答
AB中点P(x,y)
2x=xA+xB
2y=yA+yB
过Q(2,-4)作圆O:x^2+y^2=9的割线L:y+4=k(x-2)
k=(y+4)/(x-2)
割线L交圆O:x^2+y^2=9于点A,B
k=(yA-yB)/(xA-xB)=(y+4)/(x-2)
(xA)^2+(yA)^2=9.(1)
(xB)^2+(yB)^2=9.(2)
(1)-(2):
(xA-xB)*(xA+xB)+(yA-yB)*(yA+yB)=0
上方程两边除(xA-xB):
(xA+xB)+(yA+yB)*(yA-yB)/(xA-xB)=0
2x+2y*(y+4)/(x-2)=0
AB中点P的轨迹方程:
(x-1)^2+(y+2)^2=5