已知P（1,2）及圆C：X2+Y2=9,过P作两条相互垂直的弦交C于A,B,求线段AB的中点的轨迹方程

AB的中点M(x,y)
xA+xB=2xM=2x,yA+yB=2yM=2y
(xA+xB)^2=4x^2
(xA)^2+(xB)^2+2xA*xB=4x^2......(1)
(yA)^2+(yB)^2+2yA*yB=4y^2......(2)
(xA)^2+(yA)^2=9......(3)
(xB)^2+(yB)^2=9......(4)
AP⊥BP
k(AP)*k(BP)=-1
[(yA-yP)/(xA-xP)]*[(yB-yP)/(xB-xP]=-1
[(yA-2)/(xA-1)]*[(yB-2)/(xB-1]=-1
xA*xB-2(yA+yB)+4+yA*yB-(xA+xB)+1=0
2xA*xB+2yA*yB=2(xA+xB)+4(yA+yB)-10=4x+8y-10......(5)
(1)+(2)-(3)-(4)-(5):
线段AB的中点的轨迹方程:(x-0.5)^2+(y-1)^2=3.25

设A(3cosa,3sina),B(3cosb,3sinb),则中点坐标：(x,y)满足：x=3(cosa+cosb)/2,y=3(sina+sinb)/2x^2+y^2=9/4[(cosa+cosb)^2+(sina+sinb)^2]=9/4*(2+2(cosacosb+sinasinb))9(cosacosb+sinasinb)=2(x^2+y^2)-9-1=(3sina-...