Step-by-step explanation:
(x − 1)² + y² = k
y = ax
When they intersect:
(x − 1)² + (ax)² = k
x² − 2x + 1 + a²x² = k
(1 + a²) x² − 2x + 1 − k = 0
For this to have two real solutions, the discriminant must be positive.
b² − 4ac > 0
(-2)² − 4 (1 + a²) (1 − k) > 0
4 + 4 (1 + a²) (k − 1) > 0
4 (1 + a²) (k − 1) > -4
(1 + a²) (k − 1) > -1
k − 1 > -1 / (1 + a²)
k > 1 − 1 / (1 + a²)
k > (1 + a² − 1) / (1 + a²)
k > a² / (1 + a²)
