A point on hyperbola is (3secθ, 2tanθ). It lies on the circle, so, 9sec2θ+4tan2θ−24secθ = 0 ⇒ 13sec2θ−24secθ−4 = 0 ⇒ sec θ = 2 , −
2
13
Therefore, sec θ = 2 ⇒ tan θ = √3 The point of intersection are A(6,2√3) and B (6,−2√3) Hence, The circle with AB as diameter is (x−6)2+y2 = (2√3)2 ⇒ x2+y2 -12x + 24 = 0
