Question

已知直线y=mx+1与圆x^2+y^2=a总有公共点,求a的取值范围

Answer 1

x^2+(mx+1)^2=a

(m^2+1)x^2+2mx+1-a=0

delta=4m^2-4(1-a)(m^2+1)

=4m^2-4(m^2+1-am^2-a)

=4am^2+4a-4

对任意的m恒>=0

若a=0,delta=-40,则4a-4>=0 a>=1

key a>=1