A-maths!

x^2 + y^2 + 2x - 4y + 1 + k(2x + y + 4) = 0 ------- (1)

a)
代(-1,2)入(1),
搵到 k = 1

代 k = 1 入(1), 圓方程為
x^2 + y^2 + 4x - 3y + 5 = 0

b)
代 y = 0 入(1),
x^2 + 2x + 1 + k(2x + 4) = 0
x^2 + 2(1+k)x + (1+4k) = 0
4 (1+k)^2 - 4(1+4k) = 0 (相切特性)
k = 0 or k = 2
代返入(1), 搵到兩條方程答案