Question

经过点M(-6,0)作圆C:x2+y2+4x-12y+24=0的割线,交圆C于A,B两点,求线段A,B的中点P的轨迹方程.

Answer 1

C：(x+2)^2+(y-6)^2=16
C(-2,6),r=4
P(x,y)
xA+xB=2x,yA+yB=2y
k(AB)=k(PM)
(yA-yB)/(xA-xB)=y/(x+6)
(xA)^2+(yA)^2+4xA-12yA+24-[(xB)^2+(yB)^2+4xB-12yB+24]=0
(xA+xB)*(xA-xB)+(yA+yB)*(yA-yB)+4(xA-xB)-12(yA-yB)=0
(2x+4)*(xA-xB)+(2y-12)*(yA-yB)=0
(x+2)+(y-6)*(yA-yB)/(xA-xB)=0
(x+2)+(y-6)*y/(x+6)=0
(x+4)^2+(y-3)^2=13