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)`