A point on hyperbola is (3sec$$\theta$$, 2tan$$\theta$$).

It lies on the circle, so

$$9{\sec ^2}\theta + 4{\tan ^2}\theta - 24\sec \theta = 0$$.

$$ \Rightarrow 13{\sec ^2}\theta - 24\sec \theta - 4 = 0 \Rightarrow \sec \theta = 2, - {2 \over {13}}$$

Therefore, $$\sec \theta = 2 \Rightarrow \tan \theta = \sqrt 3 $$

The point of intersection are $$A(6,2\sqrt 3 )$$ and $$B(6, - 2\sqrt 3 )$$

Hence, The circle with AB as diameter is

$${(x - 6)^2} + {y^2} = {(2\sqrt 3 )^2} \Rightarrow {x^2} + {y^2} - 12x + 24 = 0$$