Question

已知圆C:x²+(y-1)²=5,直线L:mx-y+1-m=0 (1)求证:直线L与圆C总有两个不同的交点（2）设直线L与圆

交于AB两点，若绝对值AB绝对值=根号17，求L倾斜角。。。。。。。。。。。。。。。。。。。。。。不好意思的说 我不会 请人帮忙

Answer 1

圆C:x2+(y-1)2=5
直线L:mx-y+1-m=0
==>y=mx+1-m
==>x^2+(mx+1-m-1)^2-5=0
==>(m^2+1)x^2-2m^2x+m^2-5=0
==>△=4m^4-4(m^2+1)(m^2-5)
=4m^2+5
>0
==>方程有2个不等实数根
==>直线L与圆C总有两个不同的交点
2.(m^2+1)x^2-2m^2x+m^2-5=0
==>x1+x2=2m^2/(m^2+1),x1x2=(m^2-5)/(m^2+1)
==>|AB|^2=(x1-x2)^2+(y1-y2)^2
=(1+m^2)(x1-x2)^2
=(1+m^2)[(x1+x2)^2-4x1x2]
=4(4m^2+5)/(1+m^2)=17
==>m^2=3
==>m=±√3
==>L倾斜角=m=±√3