圆的方程为(k+1)x2+(k+1)y2–x–ky=0,当k≠–1时,该圆恒过两定点,则两定点的坐标分别为?
问题描述:
圆的方程为(k+1)x2+(k+1)y2–x–ky=0,当k≠–1时,该圆恒过两定点,则两定点的坐标分别为?
答
(k+1)x2+(k+1)y2–x–ky=0
k(x^2+y^2-y)+x^2+y^2-x=0
x^2+y^2-y=0
x^2+y^2-x=0
x=0,y=0
或x=1/2,y=1/2
该圆恒过两定点,则两定点的坐标分别(0,0),(1/2,1/2)