Equation of Circle on Hyperbola Intersection Chord Diameter

To find the equation of the circle with diameter $AB$, we first determine the coordinates of the points of intersection, $A$ and $B$, of the given curves:

1. Circle:
   $x^2 + y^2 - 8x = 0 \implies y^2 = 8x - x^2$
2. Hyperbola:
   $\frac{x^2}{9} - \frac{y^2}{4} = 1 \implies 4x^2 - 9y^2 = 36$

Substitute $y^2 = 8x - x^2$ into the equation of the hyperbola:
$4x^2 - 9(8x - x^2) = 36$
$4x^2 - 72x + 9x^2 = 36$
$13x^2 - 72x - 36 = 0$

Factorizing the quadratic equation:
$13x^2 - 78x + 6x - 36 = 0$
$13x(x - 6) + 6(x - 6) = 0$
$(13x + 6)(x - 6) = 0$

This yields two potential values for $x$:
$x = 6 \quad \text{or} \quad x = -\frac{6}{13}$

For real intersection points, we must have $y^2 = x(8 - x) \geq 0$, which requires $0 \leq x \leq 8$:

• If $x = -\frac{6}{13} < 0$, then $y^2 < 0$, which gives no real solutions.
• If $x = 6$, then:
  $y^2 = 8(6) - 6^2 = 48 - 36 = 12 \implies y = \pm 2\sqrt{3}$

Thus, the points of intersection are:
$A = (6, 2\sqrt{3}) \quad \text{and} \quad B = (6, -2\sqrt{3})$

The equation of a circle having the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ as its diameter is:
$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$

Substituting $A(6, 2\sqrt{3})$ and $B(6, -2\sqrt{3})$:
$(x - 6)(x - 6) + (y - 2\sqrt{3})(y + 2\sqrt{3}) = 0$
$(x - 6)^2 + (y^2 - 12) = 0$
$x^2 - 12x + 36 + y^2 - 12 = 0$
$x^2 + y^2 - 12x + 24 = 0$

Therefore, the correct option is A.