Question

与直线y=2相切,且与圆x2+(y-1)2=1外切的圆的圆心轨迹方程

Answer 1

所求圆为P(m,n)P到y=2的距离2-n即为半径P到圆x^2+(y-1)2^=1圆的距离(m^2+(n-1)^2)^0.5(m^2+(n-1)^2)^0.5=1+2-nm^2+(n-1)^2=(3-n)^2m^2+n^2-2n+1=n^2-6n+9 m^2=-4n+8n=-m^2/4+2轨迹方程y=-x^2/4+2