Question

Solve it step by step

Answer 1

case 1 :- when y = mx + 2 cut x^2 + y^2 = 1 distinct then ,
put y = mx + 2 in. x^2 +y^2 = 1

x^2 + (mx + 2)^2 = 1
x^2 + m^2x^2 + 4mx + 4 = 1
(m^2 + 1) x^2 + 4mx + 3 =0
now ,
this quadratic equation gain real roots when ,
D €[ 0, infinity)
€ means belongs to
now ,
(4m)^2 -4 (m^2+ 1) 3 >=0
4 m^2 - 3m^2 - 3 >=0
m^2 -3 >=0
(m -root3)(m +root3)>=0
hence,
m € [root3 , infinity) U (-infinity , -root3]

case 2:- when y = mx + 2 coincident of x^2 + y^2 = 1
x^2 + (mx + 2 )^2 = 1
(m^2 + 1) x ^2 + 4mx + 3 = 0
D = 0. for coincident
16m^2 = 12 (m^2 + 1)
m = +_ root3

hence ,
for case 1 and case 2 intersection of both value , gain
m €[ root3 , infinity ) U (-infinity , -root3]

Answer 2

y = m x + 2          x² + y² = 1

Point of intersection:
    x² + (mx+2)² = 1
    (1+m²) x² + 4 m x + 3 = 0
         Discriminant:  16m² - 12 - 12 m² = 4 (m² - 3)  > 0
    x is real for m ≥ √3   or   m ≤ -√3

Range of m:   (-∞, -√3]  U [√3, ∞)

   When  m = √3 or -√3   the line is a tangent to the circle.