Given curves

The circle is $x^2+y^2-8x=0$ and the hyperbola is $\frac{x^2}{9}-\frac{y^2}{4}=1.$

Let their intersection points be $A$ and $B$.

Write the circle in a useful form

From $x^2+y^2-8x=0,$ we get $x^2+y^2=8x. \qquad (1)$

Use the hyperbola equation

From $\frac{x^2}{9}-\frac{y^2}{4}=1,$ multiplying by $36$: $4x^2-9y^2=36. \qquad (2)$

Eliminate $y^2$ using (1)

From (1), $y^2=8x-x^2.$

Substitute into (2): $4x^2-9(8x-x^2)=36$ $4x^2-72x+9x^2=36$ $13x^2-72x-36=0.$

Find the $x$-coordinates of intersection points

Solve: $13x^2-72x-36=0.$

Using quadratic formula, $x=\frac{72\pm\sqrt{72^2+4\cdot 13\cdot 36}}{26}$ $=\frac{72\pm\sqrt{5184+1872}}{26}$ $=\frac{72\pm\sqrt{7056}}{26}$ $=\frac{72\pm 84}{26}.$

So, $x=6 \quad \text{or} \quad x=-\frac{6}{13}.$

Check which value gives real intersection points

Since from (1), $y^2=8x-x^2,$ for $x=6$: $y^2=48-36=12,$ so points are real.

For $x=-\frac{6}{13}$: $y^2=8\left(-\frac{6}{13}\right)-\left(-\frac{6}{13}\right)^2<0,$ which is not possible for real intersection points.

Hence the real intersection points are $A=(6,2\sqrt{3}), \qquad B=(6,-2\sqrt{3}).$

Find the circle with $AB$ as diameter

The endpoints of the diameter are $A(6,2\sqrt{3})$ and $B(6,-2\sqrt{3})$.

Their midpoint is the center: $\left(\frac{6+6}{2},\frac{2\sqrt{3}-2\sqrt{3}}{2}\right)=(6,0).$

Radius is half of $AB$: $AB=4\sqrt{3} \implies r=2\sqrt{3}.$

Therefore the required circle is $(x-6)^2+y^2=(2\sqrt{3})^2=12.$

Expanding, $x^2-12x+36+y^2=12$ $x^2+y^2-12x+24=0.$

Match with options

This is $\boxed{x^2+y^2-12x+24=0}$ which is Option A.

Comparison with stored answer

Stored correct answer: A.

Our derived answer is also A, so they agree.