设Q(x,y),A(x1,y1),B(x2,y2),而P(a,b),
有矩形性质得:x1+x2=a+x;y1+y2=b+y;
两方程平方相加得:
(x+a)^2+(y+b)^2=2r^2+2(x1x2+y1y2)
又PA⊥PB,即(b-y1)(b-y2)/(a-x1)(a-x2)=-1,
即(y2-y)(y1-y)/(x2-x)(x1-x)=-1
得x1x2+y1y2=x(x1+x2)+y(y1+y2)-x^2-y^2
=x(a+x)+y(b+y)-x^2-y^2=ax+by
所以第三个方程化为:
(x+a)^2+(y+b)^2=2r^2+2ax+2by
即x^2+y^2=2r^2-a^2-b^2