设A(x1,y1) B(x2,y2) 中点P(x,y),则有x1+x2=2x,y1+y2=2y则x1^2+y1^2=4x2^2+y2^2=4相减得(x1+x2)(x1-x2)+(y1+y2)(y1-y2)=0即k=(y1-y2)/(x1-x2)=-x/y又y=k(x-4)消去k得x^2+y^2-4x=0 （x∈[0,1]）即为中点的轨迹方程 ...