If the circles `(x-3)^(2)+(y-4)^(4)=16` and `(x-7)^(2)+y-7)^(2)=9` intersect at points A and B, then the area (in sq. units) of the quadrilateral `C_(1)AC_(2)B` is equal to (where, `C_(1) and C_(2)` are centres of the given circles)

Options

A. 6
B. 12
C. 18
D. 24

Answer

To find the area of the quadrilateral \( C_1AC_2B \), where \( C_1 \) and \( C_2 \) are the centers of the given circles, we can follow these steps:

### Step 1: Identify the centers and radii of the circles
The equations of the circles are given as:
1. \((x - 3)^2 + (y - 4)^2 = 16\)
2. \((x - 7)^2 + (y - 7)^2 = 9\)

From the first circle, we can identify:
- Center \( C_1 = (3, 4) \)
- Radius \( r_1 = \sqrt{16} = 4 \)

From the second circle, we can identify:
- Center \( C_2 = (7, 7) \)
- Radius \( r_2 = \sqrt{9} = 3 \)

### Step 2: Determine the coordinates of points A and B
Since the circles intersect at points A and B, we can find the coordinates of these points by solving the equations of the circles simultaneously. However, for the area calculation, we can use the properties of the quadrilateral formed.

### Step 3: Analyze the quadrilateral \( C_1AC_2B \)
The quadrilateral \( C_1AC_2B \) has two right triangles \( C_1AB \) and \( C_2AB \). The angle \( C_1AC_2 \) is a right angle because the radii to the points of intersection are perpendicular to the line joining the centers of the circles.

### Step 4: Calculate the area of triangle \( C_1AC_2 \)
The area of triangle \( C_1AC_2 \) can be calculated using the formula for the area of a triangle:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, we can take:
- Base \( C_1C_2 = \sqrt{(7 - 3)^2 + (7 - 4)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \)
- Height is the distance from point A (or B) to line \( C_1C_2 \). Since the triangles are right triangles, we can use the radii as the heights.

### Step 5: Calculate the area of quadrilateral \( C_1AC_2B \)
Since the area of quadrilateral \( C_1AC_2B \) is twice the area of triangle \( C_1AC_2 \):
\[ \text{Area of } C_1AC_2B = 2 \times \text{Area of } C_1AC_2 \]
Using the base and height:
- Base = \( C_1C_2 = 5 \)
- Height = \( r_1 = 4 \) (or \( r_2 = 3 \), but we can use the larger radius)

Thus:
\[ \text{Area of } C_1AC_2 = \frac{1}{2} \times 5 \times 4 = 10 \]
So, the area of the quadrilateral \( C_1AC_2B \) is:
\[ \text{Area of } C_1AC_2B = 2 \times 10 = 20 \text{ square units} \]

### Final Result
The area of the quadrilateral \( C_1AC_2B \) is \( 20 \) square units.