Question

己知直线l:y=-1和圆c=x^2+(y-2)^2=1,动圆m与l相切且与圆c外切则动圆圆心m的轨迹方程为?

Answer 1

设圆心坐标(x,y)

点到直线的距离 =|y+1|

点到圆C距离 =x^2+(y-2)^2

两距离之差等于圆C半径

根号[x^2+(y-2)^2] = (y+1+2)

x^2+(y-2)^2=(y+3)^2

x^2+y^2-4y+4=y^2+6y+9

x^2=10y+5