The circles `x^(2)+y^(2)+2x-2y+1=0` and `x^(2)+y^(2)-2x-2y+1=0` touch each other

Options

A. externally at (0, 1)
B. internally at (0, 1)
C. externally at (1, 0)
D. internally at (1, 0)

Explanation

To determine whether the circles given by the equations \(x^2 + y^2 + 2x - 2y + 1 = 0\) and \(x^2 + y^2 - 2x - 2y + 1 = 0\) touch each other, we will follow these steps:

### Step 1: Rewrite the equations in standard form
We start by rewriting the equations of the circles in standard form, which is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.

1. **For the first circle:**
\[ x^2 + y^2 + 2x - 2y + 1 = 0 \]
Rearranging gives:
\[ x^2 + 2x + y^2 - 2y + 1 = 0 \]
Completing the square:
\[ (x + 1)^2 - 1 + (y - 1)^2 - 1 + 1 = 0 \]
This simplifies to:
\[ (x + 1)^2 + (y - 1)^2 = 1 \]
Thus, the center \(C_1\) is \((-1, 1)\) and the radius \(r_1 = 1\).

2. **For the second circle:**
\[ x^2 + y^2 - 2x - 2y + 1 = 0 \]
Rearranging gives:
\[ x^2 - 2x + y^2 - 2y + 1 = 0 \]
Completing the square:
\[ (x - 1)^2 - 1 + (y - 1)^2 - 1 + 1 = 0 \]
This simplifies to:
\[ (x - 1)^2 + (y - 1)^2 = 1 \]
Thus, the center \(C_2\) is \((1, 1)\) and the radius \(r_2 = 1\).

### Step 2: Calculate the distance between the centers
Next, we calculate the distance \(d\) between the centers \(C_1\) and \(C_2\):
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of the centers:
\[ d = \sqrt{(1 - (-1))^2 + (1 - 1)^2} = \sqrt{(1 + 1)^2 + 0^2} = \sqrt{2^2} = 2 \]

### Step 3: Check the condition for touching circles
For two circles to touch externally, the distance \(d\) between their centers must equal the sum of their radii:
\[ d = r_1 + r_2 \]
Substituting the values:
\[ 2 = 1 + 1 \]
This condition is satisfied, indicating that the circles touch each other externally.

### Step 4: Find the point of tangency
The point of tangency between two externally touching circles is the midpoint of the line segment joining their centers:
\[ O = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
Substituting the coordinates of the centers:
\[ O = \left(\frac{-1 + 1}{2}, \frac{1 + 1}{2}\right) = \left(0, 1\right) \]

### Conclusion
The circles touch each other externally at the point \(O(0, 1)\).