Question

几何问题（轨迹）经过坐标原点O的直线与圆C:X²+Y²+2X-4Y-31=0相交于A,B两点,求弦AB中点M的轨迹方程.

Answer 1

设割线方程为：y=kx代人x²+y²-4y+2x+4=0得：(1+k^2)x^2+(2-4k)x+4=0x1+x2=(4k-2)/(1+k^2)所以,弦AB的中点M的横坐标=(x1+x2)/2=(2k-1)/(1+k^2)把k=y/x代人得：x=(2y/x-1)/(1+y^2/x^2)x=x(2y-x)/(x^2+y^2)x^2+y^2=2y-x所以,M的轨迹方程：x^2+y^2+x-2y=0