Equation of the circle having diameters `2x - 3y =5` and `3x - 4y =7` and radius 8 is

Options

A. `x^(2)+y^(2)-2x+2y-62=0`
B. `x^(2)+y^(2)+2x-2y+62=0`
C. `x^(2)+y^(2)+2x+2y-2=0`
D. none of these

Answer

To find the equation of the circle having diameters given by the lines \(2x - 3y = 5\) and \(3x - 4y = 7\) and a radius of 8, we can follow these steps:

### Step 1: Find the intersection of the two lines
The intersection point of the two lines will give us the center of the circle. We can solve the equations simultaneously.

1. **Equation 1:** \(2x - 3y = 5\)
2. **Equation 2:** \(3x - 4y = 7\)

To solve these equations, we can use the elimination method. We will multiply Equation 1 by 4 and Equation 2 by 3 to eliminate \(y\):

\[ 4(2x - 3y) = 4(5) \implies 8x - 12y = 20 \quad \text{(Equation 3)} \]
\[ 3(3x - 4y) = 3(7) \implies 9x - 12y = 21 \quad \text{(Equation 4)} \]

Now, we can subtract Equation 3 from Equation 4:

\[ (9x - 12y) - (8x - 12y) = 21 - 20 \]
\[ x = 1 \]

### Step 2: Substitute \(x\) back to find \(y\)
Now that we have \(x = 1\), we can substitute this value back into either of the original equations to find \(y\). Let's use Equation 1:

\[ 2(1) - 3y = 5 \]
\[ 2 - 3y = 5 \implies -3y = 5 - 2 \implies -3y = 3 \implies y = -1 \]

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

### Step 3: Write the equation of the circle
The standard form of the equation of a circle is given by:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Where \(r\) is the radius. Given that the radius \(r = 8\), we have:

\[ r^2 = 8^2 = 64 \]

Substituting \(h = 1\) and \(k = -1\) into the equation:

\[ (x - 1)^2 + (y + 1)^2 = 64 \]

### Step 4: Expand the equation
Now we will expand the equation:

\[ (x - 1)^2 + (y + 1)^2 = 64 \]
\[ (x^2 - 2x + 1) + (y^2 + 2y + 1) = 64 \]
\[ x^2 + y^2 - 2x + 2y + 2 = 64 \]
\[ x^2 + y^2 - 2x + 2y - 62 = 0 \]

### Final Answer
The equation of the circle is:

\[ x^2 + y^2 - 2x + 2y - 62 = 0 \]