Question

求助一道圆锥曲线的题目,一个动圆与x轴相切,且被直线y=x所截得的弦长为定值2,求动圆圆心的轨迹方程,一道填空题,知道怎

Answer 1

设动圆的圆心坐标是(m,n),则半径r=|n|

圆心到y=x的距离是d=|m-n|/根号2

勾股定理得:r^2=d^2+(2/2)^2

n^2=(m-n)^2/2+1

2n^2=m^2-2mn+n^2+2

m^2-2mn-n^2+2=0

即方程是x^2-2xy-y^2+2=0.