F5 MATHS

Sub y = mx+1 to (x-2)^2 + y^2 =1 :
(x-2)^2 + (mx+1)^2 = 1
(x^2 - 4x + 4) + (mx)^2 + 2mx + 1 = 1
(m^2 + 1)x^2 + 2(m - 2)x + 4 = 0
Since the staight line is tangent to the circle ,
so x have one solution only ,
△ = [2(m - 2)]^2 - 4(m^2 + 1)(4) = 0
4(m^2 - 4m + 4) - (16m^2 + 16) = 0
4m^2 - 16m + 16 - 16m^2 - 16 = 0
m^2 - 4m - 4m^2 = 0
3m^2 + 4m = 0
m(3m + 4) = 0
m = 0 or m = - 4/3