Equation of the circle which circumscribes the square formed by the lines `xy -9x - 2y + 18=0 and xy-5x - 6y + 30 =0` is

Options

A. `x ^(2) + y ^(2) - 4x - 10 y + 21 =0`
B. `x ^(2) + y ^(2) -8x - 14 y + 57 =0`
C. `x ^(2) + y ^(2) - 12 x - 10 y + 53 =0`
D. none of these

Answer

To find the equation of the circle that circumscribes the square formed by the lines \( xy - 9x - 2y + 18 = 0 \) and \( xy - 5x - 6y + 30 = 0 \), we can follow these steps:

### Step 1: Identify the lines from the equations
The given equations represent pairs of straight lines. We will factor these equations to find the lines.

1. **For the first equation**:
\[ xy - 9x - 2y + 18 = 0 \]
Group the terms:
\[ x(y - 9) - 2(y - 9) = 0 \]
Factor out \( (y - 9) \):
\[ (y - 9)(x - 2) = 0 \]
This gives us the lines:
\[ x = 2 \quad \text{and} \quad y = 9 \]

2. **For the second equation**:
\[ xy - 5x - 6y + 30 = 0 \]
Group the terms:
\[ x(y - 5) - 6(y - 5) = 0 \]
Factor out \( (y - 5) \):
\[ (y - 5)(x - 6) = 0 \]
This gives us the lines:
\[ x = 6 \quad \text{and} \quad y = 5 \]

### Step 2: Find the intersection points
The four lines \( x = 2, x = 6, y = 9, y = 5 \) will form a square. We can find the intersection points:
1. Intersection of \( x = 2 \) and \( y = 5 \): \( (2, 5) \)
2. Intersection of \( x = 2 \) and \( y = 9 \): \( (2, 9) \)
3. Intersection of \( x = 6 \) and \( y = 5 \): \( (6, 5) \)
4. Intersection of \( x = 6 \) and \( y = 9 \): \( (6, 9) \)

### Step 3: Identify the center and radius of the circle
The center of the circle is the midpoint of the diagonal formed by two opposite corners of the square. We can take the points \( (2, 5) \) and \( (6, 9) \).

1. **Midpoint**:
\[ \text{Midpoint} = \left( \frac{2 + 6}{2}, \frac{5 + 9}{2} \right) = \left( 4, 7 \right) \]

2. **Radius**:
The radius is the distance from the center to any vertex of the square. We can calculate the distance from \( (4, 7) \) to \( (2, 5) \):
\[ r = \sqrt{(4 - 2)^2 + (7 - 5)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \]

### Step 4: Write the equation of the circle
The standard form of the equation of a circle is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Where \( (h, k) \) is the center and \( r \) is the radius.

Substituting \( (h, k) = (4, 7) \) and \( r = 2\sqrt{2} \):
\[ (x - 4)^2 + (y - 7)^2 = (2\sqrt{2})^2 \]
\[ (x - 4)^2 + (y - 7)^2 = 8 \]

### Final Equation
Expanding this gives:
\[ (x^2 - 8x + 16) + (y^2 - 14y + 49) = 8 \]
Combining terms:
\[ x^2 + y^2 - 8x - 14y + 57 = 0 \]

### Conclusion
The equation of the circle that circumscribes the square is:
\[ x^2 + y^2 - 8x - 14y + 57 = 0 \]