已知A(1/2,0)为圆x^2+y^2=1内一定点,圆上有两动点P,Q恒有PA垂直QA,过P,Q做圆的两切线交于点M,求M的轨迹

题目

已知A(1/2,0)为圆x^2+y^2=1内一定点,圆上有两动点P,Q恒有PA垂直QA,过P,Q做圆的两切线交于点M,求M的轨迹
求看得懂的解答

答案

M(x,y),A(0.5,0)
(xP)^2+(yP)^2=(xQ)^2+(yQ)^2=1
PA⊥QA
[yP/(xP-0.5)]*[yQ/(xQ-0.5)]=-1
2xP*xQ+2yP*yQ=(xP+xQ)-0.5=(2x-0.5x^2+0.5y^2)/(x^2-y^2).(1)
[(y-yP)/(x-xP)]*(yP/xP)=-1
x*xP+y*yP=(xP)^2+(yP)^2=1.(2)
x*xQ+y*yQ=1.(3)
[(2)+(3)]*y:
xy*(xP+xQ)+y^2*(yP+yQ)=2y.(4)
y*(xP+xQ)+x(yP+yQ)=0.(5)
(5)*x:
xy*(xP+xQ)+x^2*(yP+yQ).(6)
(4)-(6):
(yP+yQ)=2y/(y^2-x^2)
(xP+xQ)=2x/(x^2-y^2)
(xP+xQ)^2=[2x/(x^2-y^2)]^2
2xP*xQ=4x^2/(x^2-y^2)^2-1.(7)
2yP*yQ=4y^2/(x^2-y^2)^2-1.(8)
(1)-(7)-(8):
请自己算