Question

Find the range of values of m for which the line $y=mx+2$ cuts the circle $x^2+y^2=1$ at distinct points.

Answer 1

$x^2+(mx+2{)}^2=1$

$x^2+m^2{x}^2+4mx+4=1$

$(1+m^2)x^2+4mx+3=0$

Hence, for real and unequal roots

$a\ne 0,D>0$

$1+m^2\ne 0,16m^2-12(1+m^2)>0$

or,$m^2>3$

$m>\sqrt{3}$and $m<-\sqrt{3}$