The equation of the circle passing through the point (-1, 2) and having two diameters along the pair of lines `x^(2)-y^(2)-4x+2y+3=0`, is

Options

A. `x^(2)+y^(2)-4x-2y+5=0`
B. `x^(2)+y^(2)+4x+2y-5=0`
C. `x^(2)+y^(2)-4x-2y-5=0`
D. none of these

Explanation

To find the equation of the circle that passes through the point (-1, 2) and has two diameters along the pair of lines given by the equation \(x^2 - y^2 - 4x + 2y + 3 = 0\), we can follow these steps:

### Step 1: Rewrite the given equation of the pair of lines
We start with the equation:
\[ x^2 - y^2 - 4x + 2y + 3 = 0 \]
We can rearrange this equation to group the \(x\) and \(y\) terms:
\[ (x^2 - 4x) - (y^2 - 2y) + 3 = 0 \]

### Step 2: Complete the square for \(x\) and \(y\)
To complete the square for \(x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]
For \(y\):
\[ -y^2 + 2y = -(y^2 - 2y) = -((y - 1)^2 - 1) = -(y - 1)^2 + 1 \]
Now substituting these back into the equation:
\[ ((x - 2)^2 - 4) - ((y - 1)^2 - 1) + 3 = 0 \]
This simplifies to:
\[ (x - 2)^2 - (y - 1)^2 - 4 + 1 + 3 = 0 \]
\[ (x - 2)^2 - (y - 1)^2 = 0 \]

### Step 3: Factor the equation
Using the difference of squares:
\[ (x - 2 + (y - 1))(x - 2 - (y - 1)) = 0 \]
This gives us two equations:
1. \(x + y - 3 = 0\)
2. \(x - y - 1 = 0\)

### Step 4: Find the center of the circle
To find the center of the circle, we need to find the intersection of the two lines:
1. From \(x + y - 3 = 0\), we can express \(y\) as:
\[ y = 3 - x \]
2. Substitute into the second equation:
\[ x - (3 - x) - 1 = 0 \implies 2x - 4 = 0 \implies x = 2 \]
3. Substitute \(x = 2\) back into \(y = 3 - x\):
\[ y = 3 - 2 = 1 \]
Thus, the center of the circle is \(C(2, 1)\).

### Step 5: Find the radius of the circle
The circle passes through the point (-1, 2). We can use the distance formula to find the radius:
\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting \(C(2, 1)\) and point \((-1, 2)\):
\[ r = \sqrt{((-1) - 2)^2 + (2 - 1)^2} = \sqrt{(-3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10} \]

### Step 6: Write the equation of the circle
The standard form of the equation of a circle is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = 2\), \(k = 1\), and \(r^2 = 10\):
\[ (x - 2)^2 + (y - 1)^2 = 10 \]

### Step 7: Expand the equation
Expanding the equation:
\[ (x^2 - 4x + 4) + (y^2 - 2y + 1) = 10 \]
Combining like terms:
\[ x^2 + y^2 - 4x - 2y + 5 - 10 = 0 \]
This simplifies to:
\[ x^2 + y^2 - 4x - 2y - 5 = 0 \]

### Final Answer
The equation of the circle is:
\[ \boxed{x^2 + y^2 - 4x - 2y - 5 = 0} \]