Question

A circle has equation $(x-1)^{2}+y^{2}=k$ , where k > 0.
The straight line L with the equation $y=ax$ cuts the circle at two distinct points.
Prove that $k\ \textgreater \ \frac{a^{2} }{1+a^{2} }$

Answer 1

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)