Question 985482
{{{ system(x^2 + y^2 = 4, y = mx + 4) }}} gives us the intersection points of circle and line.
{{{ system(x^2 + y^2 = 4, y = mx + 4) }}}--->{{{x^2 + (mx + 4)^2 = 4}}}--->{{{x^2 + m^2x^2 +8mx+16 = 4}}}--->{{{(m^2+1)x^2 +8mx+12 =0}}}
That is a quadratic equation.
A quadratic equation of the form {{{ax^2+bx+c=0}}}
has no real solutions if {{{b^2-4*a*c<0}}} .
It has one real solution if {{{b^2-4*a*c=0}}} , and
it has two real solutions if {{{b^2-4*a*c>0}}} .
For the quadratic equation {{{(m^2+1)x^2 +8mx+12 =0}}} , {{{system(a=m^2+1,b=8m,c=12)}}} ,so
{{{b^2-4*a*c=(8m)^2-4(m^2+1)*12=64m^2-48m^2-48=16m^2-48=16(m^2-3)}}} .
The zeros of {{{16(m^2-3)}}} are the zeros of {{{m^2-3}}} , given by
{{{m=-sqrt(3)}}} and {{{m=sqrt(3)}}} .
For those values, the line is tangent to the circle.
{{{drawing(300,300,-5,5,-4,6,
grid(0),circle(0,0,2),
line(-5,-4.66,5,12.66),
line(-5,12.66,5,-4.66)
)}}}
In between those values of {{{m}}} ,
for {{{-sqrt(3)<m<sqrt(3)}}} ,
{{{16(m^2-3)}}} is negative, and the line does not intersect the circle.
Otherwise, for {{{system(m<-sqrt(3),"or",m>sqrt(3))}}} ,
{{{16(m^2-3)>0)}}} , and the line intersects the circle  in two places.