The circle $x^{2}+y^{2}-8x=0$ and hyperbola $\dfrac{x^{2}}{9}-\dfrac{y^{2}}{4}=1$ intersect at the points $A$ and $B$.

then the equation of the circle with $AB$ as its diameter is

Given: $x^2+y^2-8x=0 .....(1)$ $\implies y^2=8x-x^2$

$\dfrac {x^2}{9}-\dfrac {y^2}{4}\Rightarrow 1$

$\Rightarrow\dfrac {x^2}{9}-\dfrac {(8x-x^2)}{4}=1$

$\Rightarrow4x^2-9(8x-x^2)=36$

$\Rightarrow4x^2-72x+9x^2-36=0$

$\Rightarrow13x^2-72x-36=0$

$\Rightarrow x=\dfrac {72\pm \sqrt {(72)^2-4\times 13\times 36}}{26}$

$\Rightarrow x=6$

Then, putting $x=6$ in equation (1), we get

$\Rightarrow y=\pm \sqrt{12}=\pm \sqrt{4\times3}=\pm 2\sqrt 3$

So, radius $=2\sqrt 3$

As $AB$ is a diameter center will be the midpoint of point $A=(6,2\sqrt 3)$ and

$B=(6,-2\sqrt 3)$

By mid point formula,

Center $=\left(\dfrac{6+6}{2},\dfrac{2\sqrt 3-2\sqrt 3}{2}\right)$

Centre $=(6, 0)$

$\Rightarrow(x-6)^2+y^2=12$

$\Rightarrow x^2+36-12x+y^2=12$

$\Rightarrow x^2-12x+y^2+24=0$