已知圆C的方程为:x^2+y^2=9,圆内定点P(1,-1),圆周上有两个动点A,B,使PA⊥PB,求矩形 急

Q(,x,y)
xA+xB=xQ+xP=x+1,yA+yB=yQ+yP=y-1
(xA+xB)^2=(x+1)^2
(xA)^2+(xB)^2+2xA*xB=(x+1)^2.(1)
(yA)^2+(yB)^2+2yA*yB=(y-1)^2.(2)
PA⊥PB,k(PA)*k(PB)=-1
[(yA+1)/(xA-1)]*[(yB+1)/(xB-1)]=-1
xA*xB+yA*yB=(xA+xB)-(yA+yB-2=x+1-(y-1)-2=x-y
2(xA*xB+yA*yB)=2x-2y.(3)
(1)+(2)-(3):
(xA)^2+(yA)^2+(xB)^2+(yB)^2=(x+1)^2+(y-1)^2-(2x-2y)
9+9=(x+1)^2+(y-1)^2-(2x-2y)
矩形APBQ的顶点Q的轨迹方程:x^2+(y-2)^2=20