Question

过圆x^2+y^2=r^2外一点M（a,b)作圆的割线,求所得弦中点的轨迹方程

Answer 1

AB中点P(x,y)

xA+xB=2x

yA+yB=2y

(xA)^2+(yA)^2=r^2.(1)

(xB)^2+(yB)=4y^2.(2)

(1)-(2):

(xA+xB)*(xA-xB)+(yA+yB)*(yA-yB)=0

(xA+xB) +( yA+yB)*(yA-yB)/(xA-xB)=0

k(AB)=(yA-yB)/(xA-xB)=(y-b)/(x-a)

2x+2y*(y-b)/(x-a)=0

AB中点的轨迹方程是圆:x^2+y^2-ax-by=0