For the given circles `x^(2)+y^(2)-6x-2y+1=0` and `x^(2)+y^(2)+2x-8y+13=0`, which of the following is true?

Options

A. One circle lies inside the other
B. One circle lies completely outside the other
C. Two circles intersection in two points
D. They touch each other externally

Explanation

To solve the problem regarding the given circles, we will follow these steps:

### Step 1: Write the equations of the circles
The equations of the circles are given as:
1. Circle 1: \( x^2 + y^2 - 6x - 2y + 1 = 0 \)
2. Circle 2: \( x^2 + y^2 + 2x - 8y + 13 = 0 \)

### Step 2: Identify the center and radius of Circle 1
The general form of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From Circle 1, we can identify:
- \( 2g = -6 \) → \( g = -3 \)
- \( 2f = -2 \) → \( f = -1 \)
- \( c = 1 \)

The center (C1) and radius (R1) can be calculated as follows:
- Center \( C1 = (-g, -f) = (3, 1) \)
- Radius \( R1 = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + (-1)^2 - 1} = \sqrt{9 + 1 - 1} = \sqrt{9} = 3 \)

### Step 3: Identify the center and radius of Circle 2
For Circle 2, we have:
- \( 2g = 2 \) → \( g = 1 \)
- \( 2f = -8 \) → \( f = -4 \)
- \( c = 13 \)

The center (C2) and radius (R2) are:
- Center \( C2 = (-g, -f) = (-1, 4) \)
- Radius \( R2 = \sqrt{g^2 + f^2 - c} = \sqrt{(1)^2 + (-4)^2 - 13} = \sqrt{1 + 16 - 13} = \sqrt{4} = 2 \)

### Step 4: Calculate the distance between the centers
The distance \( d \) between the centers \( C1(3, 1) \) and \( C2(-1, 4) \) is calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{((-1) - 3)^2 + (4 - 1)^2} = \sqrt{(-4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Step 5: Analyze the relationship between the circles
Now we compare the distance \( d \) with the sum of the radii \( R1 + R2 \):
- \( R1 + R2 = 3 + 2 = 5 \)
- Distance \( d = 5 \)

Since \( R1 + R2 = d \), the circles are touching each other externally.

### Conclusion
The correct statement regarding the two circles is that they are touching each other externally.