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

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