The circle `x^2+y^2-8x = 0` and hyperbola `x^2 /9 - y^2 /4=1` intersect at the points A and B. Then the equation of the circle with AB as its diameter is

Options

A. a. `x^(2)+y^(2)-12x+24=0`
B. b. `x^(2)+y^(2)+12x+24=0`
C. c. `x^(2)+y^(2)+24x-12=0`
D. d. `x^(2)+y^(2)-24x-12=0`

Explanation

To solve the problem, we need to find the intersection points of the given circle and hyperbola, and then determine the equation of the circle with those points as its diameter.

### Step 1: Write the equations of the circle and hyperbola
The equations given are:
1. Circle: \( x^2 + y^2 - 8x = 0 \)
2. Hyperbola: \( \frac{x^2}{9} - \frac{y^2}{4} = 1 \)

### Step 2: Rewrite the equation of the circle
We can rewrite the circle's equation in standard form:
\[ x^2 - 8x + y^2 = 0 \implies (x - 4)^2 + y^2 = 16 \]
This represents a circle centered at (4, 0) with a radius of 4.

### Step 3: Substitute \(y^2\) from the circle into the hyperbola
From the circle, we have:
\[ y^2 = 8x - x^2 \]
Now substitute \(y^2\) into the hyperbola's equation:
\[ \frac{x^2}{9} - \frac{(8x - x^2)}{4} = 1 \]

### Step 4: Simplify the equation
Multiply through by 36 (the least common multiple of 9 and 4) to eliminate the denominators:
\[ 4x^2 - 9(8x - x^2) = 36 \]
Expanding this gives:
\[ 4x^2 - 72x + 9x^2 = 36 \]
Combine like terms:
\[ 13x^2 - 72x - 36 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 13\), \(b = -72\), and \(c = -36\).
\[ x = \frac{72 \pm \sqrt{(-72)^2 - 4 \cdot 13 \cdot (-36)}}{2 \cdot 13} \]
Calculating the discriminant:
\[ = 5184 + 1872 = 7056 \]
Now calculate \(x\):
\[ x = \frac{72 \pm \sqrt{7056}}{26} \]
\[ \sqrt{7056} = 84 \implies x = \frac{72 \pm 84}{26} \]
Calculating the two values:
1. \(x = \frac{156}{26} = 6\)
2. \(x = \frac{-12}{26} = -\frac{6}{13}\)

### Step 6: Find corresponding \(y\) values
Using \(y^2 = 8x - x^2\):
1. For \(x = 6\):
\[ y^2 = 8(6) - 6^2 = 48 - 36 = 12 \implies y = \pm 2\sqrt{3} \]
So, points \(A(6, 2\sqrt{3})\) and \(B(6, -2\sqrt{3})\).

### Step 7: Find the center and radius of the circle with AB as diameter
The center \(C\) of the circle is the midpoint of \(A\) and \(B\):
\[ C\left(\frac{6 + 6}{2}, \frac{2\sqrt{3} + (-2\sqrt{3})}{2}\right) = (6, 0) \]
The distance \(AB\) (diameter) is:
\[ AB = \sqrt{(6 - 6)^2 + (2\sqrt{3} - (-2\sqrt{3}))^2} = \sqrt{0 + (4\sqrt{3})^2} = 4\sqrt{3} \]
Thus, the radius \(r\) is:
\[ r = \frac{4\sqrt{3}}{2} = 2\sqrt{3} \]

### Step 8: Write the equation of the circle
The equation of the circle in standard form is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = 6\), \(k = 0\), and \(r = 2\sqrt{3}\):
\[ (x - 6)^2 + (y - 0)^2 = (2\sqrt{3})^2 \]
\[ (x - 6)^2 + y^2 = 12 \]
Expanding this gives:
\[ x^2 - 12x + 36 + y^2 = 12 \implies x^2 + y^2 - 12x + 24 = 0 \]

### Final Answer
The equation of the circle with AB as its diameter is:
\[ x^2 + y^2 - 12x + 24 = 0 \]