The lines 2x – 3y = 5 and 3x – 4y = 7 are diameters of a circle having area as 154 sq. units. Then the equation of the circle is

Options

A. `x^(2) + y^(2) - 2x + 2y = 62`
B. `x^(2) + y^(2) + 2x - 2y = 62`
C. `x^(2) + y^(2) + 2x - 2y = 47`
D. `x^(2) + y^(2) - 2x + 2y = 47`

Answer

To find the equation of the circle given that the lines \(2x - 3y = 5\) and \(3x - 4y = 7\) are diameters of the circle and the area of the circle is \(154\) square units, we can follow these steps:

### Step 1: Find the Center of the Circle
The center of the circle can be found by solving the equations of the two diameters.

1. Rewrite the equations of the diameters:
- \(2x - 3y - 5 = 0\) (Equation 1)
- \(3x - 4y - 7 = 0\) (Equation 2)

2. Solve these two equations simultaneously to find the point of intersection, which will be the center \((h, k)\) of the circle.

**Multiply Equation 1 by 3:**
\[ 6x - 9y - 15 = 0 \quad \text{(Equation 3)} \]

**Multiply Equation 2 by 2:**
\[ 6x - 8y - 14 = 0 \quad \text{(Equation 4)} \]

3. Subtract Equation 4 from Equation 3:
\[ (6x - 9y - 15) - (6x - 8y - 14) = 0 \]
\[ -y - 1 = 0 \Rightarrow y = -1 \]

4. Substitute \(y = -1\) back into Equation 1 to find \(x\):
\[ 2x - 3(-1) = 5 \]
\[ 2x + 3 = 5 \Rightarrow 2x = 2 \Rightarrow x = 1 \]

Thus, the center of the circle is \((1, -1)\).

### Step 2: Find the Radius of the Circle
The area of the circle is given by the formula:
\[ \text{Area} = \pi r^2 \]
Given that the area is \(154\) square units:
\[ \pi r^2 = 154 \]
Using \(\pi \approx \frac{22}{7}\):
\[ \frac{22}{7} r^2 = 154 \]
Multiplying both sides by \(7\):
\[ 22 r^2 = 1078 \]
Dividing by \(22\):
\[ r^2 = \frac{1078}{22} = 49 \Rightarrow r = 7 \]

### Step 3: Write the Equation of the Circle
The general equation of a circle with center \((h, k)\) and radius \(r\) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = 1\), \(k = -1\), and \(r^2 = 49\):
\[ (x - 1)^2 + (y + 1)^2 = 49 \]

### Step 4: Expand the Equation
Expanding the equation:
\[ (x - 1)^2 + (y + 1)^2 = 49 \]
\[ x^2 - 2x + 1 + y^2 + 2y + 1 = 49 \]
\[ x^2 + y^2 - 2x + 2y + 2 - 49 = 0 \]
\[ x^2 + y^2 - 2x + 2y - 47 = 0 \]

### Final Equation of the Circle
The equation of the circle is:
\[ x^2 + y^2 - 2x + 2y - 47 = 0 \]

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