For the circle #x^2+y^2=4# and the line #y=mx+4#, determine the exact values of the gradient, #m#, so that the line: #a)# is a tangent to the circle? #b)# intersects the circle in 2 places? #c)# does not intersect the circle?

#"substitute "y=mx+4" into the equation of the circle"#

#rArrx^2+(mx+4)^2=4larr" expand and simplify"#

#rArrx^2+m^2x^2+8mx+16-4=0#

#rArr(1+m^2)x^2+8mx+12=0#

#"with "a=1+m^2,b=8m,c=12#

#"utilise the coditions for the "color(blue)"discriminant"#

#Delta=b^2-4ac#

#• " if "Delta>0" then 2 real solutions"to(b)" above"#

#• " if "Delta=0" then 1 real solution "to(a)" above"#

#• " if "Delta<0" then no real solutions "to(c)" above"#

#Delta=b^2-4ac=64m^2-48(1+m^2)#

#color(white)(xxxxxxxx)=16m^2-48#

#(a)#

#"solve "16m^2-48=0#

#rArr16(m^2-3)=0#

#rArrm^2-3=0rArrm^2=3rArrm=+-sqrt3#

#"there are 2 possible tangents to the circle"#

#y=-sqrt3x+4" or "y=sqrt3x+4#

#(b)" and "(c)#

#"we require to solve "#

#m^2-3>0" and "m^2-3<0#

#"graphing "m^2-3" will provide both solutions"#

#"zeros are "m=+-sqrt3#

#"coefficient of "m^2>0rArruuu#
graph{x^2-3 [-10, 10, -5, 5]}

#"solutions to "m^2-3>0" lie above the m(x)-axis"#

#rArrm<-sqrt3" or "m>sqrt3#

#"in interval notation "(-oo,-sqrt3)uu(sqrt3,+oo)#

#"solutions to "m^2-3<0" lie below the m(x)-axis"#

#rArr-sqrt3< m < sqrt3" or "(-sqrt3,sqrt3)#