Let `C_(1) and C_(2) ` be the circles `x^(2) + y^(2) - 2x - 2y - 2 = 0` and `x^(2) + y^(2) - 6x - 6y + 14 = 0 ` respectively. If P and Q are points of intersection of these circles, then the area (in sq. units ) of the quadrilateral `PC_(1) QC_(2)` is :

Options

A. 4
B. 9
C. 6
D. 8

Answer

To find the area of the quadrilateral \( PC_1 QC_2 \) formed by the points of intersection \( P \) and \( Q \) of the circles \( C_1 \) and \( C_2 \), we will follow these steps:

### Step 1: Rewrite the equations of the circles in standard form

The equations of the circles are given as:
1. \( C_1: x^2 + y^2 - 2x - 2y - 2 = 0 \)
2. \( C_2: x^2 + y^2 - 6x - 6y + 14 = 0 \)

We will complete the square for both circles.

**For Circle \( C_1 \):**
\[ x^2 - 2x + y^2 - 2y = 2 \]
Completing the square:
\[ (x-1)^2 - 1 + (y-1)^2 - 1 = 2 \]
\[ (x-1)^2 + (y-1)^2 = 4 \]
Thus, the center of \( C_1 \) is \( (1, 1) \) and the radius is \( 2 \).

**For Circle \( C_2 \):**
\[ x^2 - 6x + y^2 - 6y = -14 \]
Completing the square:
\[ (x-3)^2 - 9 + (y-3)^2 - 9 = -14 \]
\[ (x-3)^2 + (y-3)^2 = 4 \]
Thus, the center of \( C_2 \) is \( (3, 3) \) and the radius is \( 2 \).

### Step 2: Find the distance between the centers of the circles

The distance \( d \) between the centers \( C_1(1, 1) \) and \( C_2(3, 3) \) is calculated using the distance formula:
\[ d = \sqrt{(3-1)^2 + (3-1)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \]

### Step 3: Determine the area of the quadrilateral \( PC_1 QC_2 \)

The quadrilateral \( PC_1 QC_2 \) can be divided into two triangles \( PC_1C_2 \) and \( QC_1C_2 \). Since both triangles are congruent (as they share the same base \( C_1C_2 \) and have the same height from points \( P \) and \( Q \)), we can calculate the area of one triangle and then double it.

The area \( A \) of triangle \( PC_1C_2 \) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base \( C_1C_2 = d = 2\sqrt{2} \) and the height from point \( P \) (or \( Q \)) to line \( C_1C_2 \) is equal to the radius of the circles, which is \( 2 \).

Thus, the area of triangle \( PC_1C_2 \) is:
\[ A = \frac{1}{2} \times 2\sqrt{2} \times 2 = 2\sqrt{2} \]

### Step 4: Calculate the area of the quadrilateral \( PC_1 QC_2 \)

Since the area of quadrilateral \( PC_1 QC_2 \) is twice the area of triangle \( PC_1C_2 \):
\[ \text{Area of } PC_1 QC_2 = 2 \times A = 2 \times 2\sqrt{2} = 4\sqrt{2} \]

### Final Answer:
The area of the quadrilateral \( PC_1 QC_2 \) is \( 4 \) square units.

---