圆x^2+y^2=1,求点[(x(x+y),y(x+y)]的轨迹方程

令m=x(x+y),n=y(x+y)则m-n=x^2-y^2,于是m-n+1=2x^2,1-m+n=2y^2(m+n-1)^2=4x^2y^2=(m-n+1)(1-m+n)整理得：m^2+n^2-m-n=0于是点[(x(x+y),y(x+y)]的轨迹方程为：x^2+y^2-x-y=0