Question

已知圆x2+y2=16和C(5,0),A,B为圆周上两个动点,满足角ACB=90°,求弦AB中点P的轨迹

Answer 1

P(x,y)

xA+xB=2x,yA+yB=2y

(xA)^2+(xB)^2+2xA*xB=4x^2.(1)

(yA)^2+(yB)^2+2yA*yB=4y^2.(2)

(xA)^2+(yA)^2=16.(3)

(xB)^2+(yB)^2=16.(4)

k(AC)*k(BC)=-1

[yA/(xA-5)]*[yB/(xB-5)]=-1

2xA*xB+2yA*yB=20x-50.(5)

(1)+(2)-(3)-(4)-(5)

(x-2.5)^2+y^2=1.75