If the lines 3x + y = 11 and x - y = 1 are the diameters of a circle of area 154 sq. units, then the equation of the circle is

Options

A. `(x-3)^(2) + (y-2)^(2) = (7)^(2)`
B. `(x+3)^(2) + (y+2)^(2) = (7)^(2)`
C. `(x-2)^(2) + (y-3)^(2) = (7)^(2)`
D. `(x+2)^(2) + (y+3)^(2) = (7)^(2)`

Answer

To find the equation of the circle given the lines \(3x + y = 11\) and \(x - y = 1\) as diameters and an area of 154 square units, we can follow these steps:

### Step 1: Find the intersection point of the lines
The intersection point of the lines will be the center of the circle. We need to solve the equations of the lines simultaneously.

1. The equations of the lines are:
\[ 3x + y = 11 \quad (1) \]
\[ x - y = 1 \quad (2) \]

2. From equation (2), we can express \(y\) in terms of \(x\):
\[ y = x - 1 \]

3. Substitute \(y\) in equation (1):
\[ 3x + (x - 1) = 11 \]
Simplifying this gives:
\[ 4x - 1 = 11 \]
\[ 4x = 12 \implies x = 3 \]

4. Substitute \(x = 3\) back into equation (2) to find \(y\):
\[ y = 3 - 1 = 2 \]

Thus, the center of the circle is \((3, 2)\).

### Step 2: Calculate the radius from the area
We know the area of the circle is given by:
\[ \text{Area} = \pi r^2 \]

Given that the area is 154 square units, we can set up the equation:
\[ \pi r^2 = 154 \]

Using \(\pi \approx \frac{22}{7}\):
\[ \frac{22}{7} r^2 = 154 \]

To eliminate the fraction, multiply both sides by 7:
\[ 22 r^2 = 154 \times 7 \]
Calculating \(154 \times 7\):
\[ 154 \times 7 = 1078 \]
So, we have:
\[ 22 r^2 = 1078 \]

Now, divide both sides by 22:
\[ r^2 = \frac{1078}{22} = 49 \]

Taking the square root gives:
\[ r = 7 \]

### Step 3: Write the equation of the circle
The general equation of a circle with center \((h, k)\) and radius \(r\) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]

Substituting \(h = 3\), \(k = 2\), and \(r = 7\):
\[ (x - 3)^2 + (y - 2)^2 = 7^2 \]
\[ (x - 3)^2 + (y - 2)^2 = 49 \]

### Final Answer
The equation of the circle is:
\[ (x - 3)^2 + (y - 2)^2 = 49 \]