Question

已知圆的方程是x^2+y^2+2(m-1)x-4my+5m^2-2m-8=0,则此圆的圆心轨迹方程是

Answer 1

(X+(m-1))^2+(Y-2m)^2=-(5m^2-2m-8)-(m^2-2m+1)-4m^2

圆心为[-(m-1),2m]

因为求轨迹,设圆心坐标(x,y);

x=-m+1

y=2m

化简得y=2(1-X)

即2X+y-2=0

如果配方没错那就应该没错了