Verified answer

A point on hyperbola is (3secΘ, 2tanΘ)

It lies on the circle, so 9sec²Θ + 4tan²Θ − 24secΘ = 0

⇒ 13sec²Θ - 24secΘ - 4 = 0

⇒ secΘ = 2, -2/13

∴ secΘ = 2

⇒ tanΘ = √3

The point of intersection are A(6, 2√3) and B(6, − 2√3)

∴ The circle with AB as diameter is

(x - 6)² + y² = (2√3)²

⇒ x² + y² - 12x + 24 = 0

Hope this helps.