A rhombus is inscribed in the region common to the two circles `x^2+y^2-4x-12=0` and `x^2+y^2+4x-12=0` with two of its vertices on the line joining the centers of the circles. The area of the rhombus is (a)`8sqrt(3)s qdotu n i t s` (b) `4sqrt(3)s qdotu n i t s` `6sqrt(3)s qdotu n i t s` (d) none of these

Options

A. `8 sqrt(3)` sq. units
B. `4 sqrt(3)` sq. units
C. `6 sqrt(3)` sq. units
D. none of these

Explanation

To find the area of the rhombus inscribed in the region common to the two circles, we will follow these steps:

### Step 1: Identify the equations of the circles
The given equations of the circles are:
1. \( x^2 + y^2 - 4x - 12 = 0 \)
2. \( x^2 + y^2 + 4x - 12 = 0 \)

### Step 2: Rewrite the equations in standard form
We will complete the square for both equations.

**For the first circle:**
\[ x^2 - 4x + y^2 - 12 = 0 \]
Completing the square for \(x\):
\[ (x - 2)^2 - 4 + y^2 - 12 = 0 \implies (x - 2)^2 + y^2 = 16 \]
This gives us the center \((2, 0)\) and radius \(4\).

**For the second circle:**
\[ x^2 + 4x + y^2 - 12 = 0 \]
Completing the square for \(x\):
\[ (x + 2)^2 - 4 + y^2 - 12 = 0 \implies (x + 2)^2 + y^2 = 16 \]
This gives us the center \((-2, 0)\) and radius \(4\).

### Step 3: Determine the line joining the centers
The centers of the circles are at \((2, 0)\) and \((-2, 0)\). The line joining these centers is the x-axis (y = 0).

### Step 4: Find the points of intersection of the circles
To find the region common to both circles, we can set the equations equal to each other:
\[ (x - 2)^2 + y^2 = 16 \quad \text{and} \quad (x + 2)^2 + y^2 = 16 \]
Subtracting these two equations:
\[ (x - 2)^2 - (x + 2)^2 = 0 \]
Expanding:
\[ (x^2 - 4x + 4) - (x^2 + 4x + 4) = 0 \implies -8x = 0 \implies x = 0 \]
Substituting \(x = 0\) into either circle's equation:
\[ 0^2 + y^2 = 16 \implies y^2 = 16 \implies y = \pm 4 \]
Thus, the points of intersection are \((0, 4)\) and \((0, -4)\).

### Step 5: Determine the rhombus vertices
The rhombus has two vertices on the line joining the centers (x-axis) and two vertices at the intersection points. Therefore, the vertices of the rhombus are:
- \(A(0, 4)\)
- \(B(0, -4)\)
- \(C(2, 0)\)
- \(D(-2, 0)\)

### Step 6: Calculate the area of the rhombus
The area of a rhombus can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times d_1 \times d_2 \]
where \(d_1\) and \(d_2\) are the lengths of the diagonals.

- The length of diagonal \(AC\) (vertical) is \(4 - (-4) = 8\).
- The length of diagonal \(BD\) (horizontal) is \(2 - (-2) = 4\).

Thus, the area is:
\[ \text{Area} = \frac{1}{2} \times 8 \times 4 = 16 \]

### Step 7: Calculate the area of the triangles within the rhombus
The rhombus can also be divided into two triangles. Each triangle has a base of \(4\) (the distance between \(C\) and \(D\)) and height of \(4\) (the distance from the x-axis to either \(A\) or \(B\)):
\[ \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8 \]
Since there are two triangles:
\[ \text{Total Area} = 2 \times 8 = 16 \]

### Conclusion
The area of the rhombus is \(16\) square units, which does not match any of the provided options. Therefore, the answer is (d) none of these.