Equation of circle of area 154 sq. units , two of whose
diameters are ` 2x - 3y + 12 = 0 " and " x + 4y - 5 = 0 ` , is

Options

A. `x^(2) + y^(2) - 6x - 4y + 36 = 0 `
B. `x^(2) + y^(2) + 6x + 4y - 36 = 0 `
C. `x^(2) + y^(2) + 6x - 4y - 36 = 0 `
D. `x^(2) + y^(2) - 6x + 4y + 36 = 0 `

Answer

To find the equation of the circle with an area of 154 square units and two diameters given by the equations \(2x - 3y + 12 = 0\) and \(x + 4y - 5 = 0\), we can follow these steps:

### Step 1: Find the center of the circle
The center of the circle is the point of intersection of the two diameters. We will solve the equations simultaneously.

1. The first equation is:
\[ 2x - 3y + 12 = 0 \quad \text{(Equation 1)} \]
2. The second equation is:
\[ x + 4y - 5 = 0 \quad \text{(Equation 2)} \]

To eliminate \(x\), we can multiply Equation 2 by 2:
\[ 2(x + 4y - 5) = 0 \implies 2x + 8y - 10 = 0 \quad \text{(Equation 3)} \]

Now, we subtract Equation 1 from Equation 3:
\[ (2x + 8y - 10) - (2x - 3y + 12) = 0 \]
This simplifies to:
\[ 11y - 22 = 0 \implies 11y = 22 \implies y = 2 \]

Now, substitute \(y = 2\) back into Equation 2 to find \(x\):
\[ x + 4(2) - 5 = 0 \implies x + 8 - 5 = 0 \implies x + 3 = 0 \implies x = -3 \]

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

### Step 2: Find the radius of the circle
The area of the circle is given as \(154\) square units. The area \(A\) of a circle is given by the formula:
\[ A = \pi r^2 \]
Using \(\pi \approx \frac{22}{7}\), we can set up the equation:
\[ 154 = \frac{22}{7} r^2 \]

To find \(r^2\), we rearrange the equation:
\[ r^2 = \frac{154 \times 7}{22} = \frac{1078}{22} = 49 \]
Taking the square root gives:
\[ r = \sqrt{49} = 7 \text{ units} \]

### Step 3: Write the equation of the circle
The standard form of the equation of a circle with center \((h, k)\) and radius \(r\) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = -3\), \(k = 2\), and \(r = 7\):
\[ (x + 3)^2 + (y - 2)^2 = 49 \]

### Step 4: Expand the equation
Expanding the equation:
\[ (x + 3)^2 + (y - 2)^2 = 49 \]
\[ x^2 + 6x + 9 + y^2 - 4y + 4 = 49 \]
Combining like terms:
\[ x^2 + y^2 + 6x - 4y + 13 - 49 = 0 \]
\[ x^2 + y^2 + 6x - 4y - 36 = 0 \]

Thus, the final equation of the circle is:
\[ x^2 + y^2 + 6x - 4y - 36 = 0 \]