The equation of one of the circles which touch the pair of lines `x^2 - y^2 + 2y - 1 = 0` is

Options

A. `x^(2) + y^(2) + 2x + 1 = 0`
B. `x^(2) + y^(2) - 2x + 1 = 0 `
C. `x^(2) + y^(2) + 2y -1 =0`
D. `x^(2) + y^(2) - 2y - 1 = 0`

Answer

To find the equation of one of the circles that touches the pair of lines given by the equation \( x^2 - y^2 + 2y - 1 = 0 \), we can follow these steps:

### Step 1: Rewrite the given equation
The given equation is:
\[ x^2 - y^2 + 2y - 1 = 0 \]
We can rearrange it to isolate \(y\):
\[ x^2 = y^2 - 2y + 1 \]
This simplifies to:
\[ x^2 = (y - 1)^2 \]

### Step 2: Identify the lines
From the equation \(x^2 = (y - 1)^2\), we can take the square root of both sides:
\[ y - 1 = \pm x \]
This gives us two lines:
1. \(y = x + 1\)
2. \(y = -x + 1\)

### Step 3: Determine the angle bisector
The two lines intersect at the point \((0, 1)\). The angle bisector of these lines is the y-axis (since the slopes of the lines are 1 and -1).

### Step 4: Find the centers of the circles
The centers of the circles that touch these lines will lie on the angle bisector (y-axis). The possible centers can be:
1. \( (0, r) \) where \(r\) is the radius of the circle.
2. \( (0, -r) \)

### Step 5: Determine the radius
Since the circle must touch both lines, we can find the radius \(r\) by calculating the distance from the center to one of the lines. The distance from the point \((0, r)\) to the line \(y = x + 1\) can be calculated using the formula for the distance from a point to a line.

The line can be rewritten in the form \(Ax + By + C = 0\):
\[ x - y + 1 = 0 \quad \text{(where } A = 1, B = -1, C = 1\text{)} \]
The distance \(D\) from the point \((0, r)\) to the line is given by:
\[ D = \frac{|A \cdot 0 + B \cdot r + C|}{\sqrt{A^2 + B^2}} = \frac{|0 - r + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|1 - r|}{\sqrt{2}} \]

### Step 6: Set the distance equal to the radius
For the circle to touch the line, the distance \(D\) must equal the radius \(r\):
\[ r = \frac{|1 - r|}{\sqrt{2}} \]

### Step 7: Solve for \(r\)
Squaring both sides:
\[ r^2 = \frac{(1 - r)^2}{2} \]
Multiplying both sides by 2:
\[ 2r^2 = (1 - r)^2 \]
Expanding the right side:
\[ 2r^2 = 1 - 2r + r^2 \]
Rearranging gives:
\[ r^2 + 2r - 1 = 0 \]
Using the quadratic formula:
\[ r = \frac{-2 \pm \sqrt{4 + 4}}{2} = -1 \pm \sqrt{2} \]
Since \(r\) must be positive, we take:
\[ r = -1 + \sqrt{2} \]

### Step 8: Write the equation of the circle
The equation of the circle with center \((0, -1 + \sqrt{2})\) and radius \(r\) is:
\[ x^2 + \left(y - (-1 + \sqrt{2})\right)^2 = (-1 + \sqrt{2})^2 \]
Simplifying gives:
\[ x^2 + \left(y + 1 - \sqrt{2}\right)^2 = 3 - 2\sqrt{2} \]

### Conclusion
The equation of one of the circles that touches the pair of lines is:
\[ x^2 + \left(y + 1 - \sqrt{2}\right)^2 = 3 - 2\sqrt{2} \]