The circles `x^(2)+y^(2)-2x-4y+1=0` and `x^(2)+y^(2)+4y+4x-1 =0`

Options

A. touch internally
B. touch externally
C. have `3x+4y-1=0` as the common tangent at the point of contact.
D. have `3x+4y+1=0` as the common tangent at the point of contact.

Explanation

To solve the problem of finding the relationship between the two circles given by the equations \(x^2 + y^2 - 2x - 4y + 1 = 0\) and \(x^2 + y^2 + 4y + 4x - 1 = 0\), we will follow these steps:

### Step 1: Rewrite the equations of the circles in standard form

1. **Circle 1**:
\[ x^2 + y^2 - 2x - 4y + 1 = 0 \]
Rearranging gives:
\[ (x^2 - 2x) + (y^2 - 4y) = -1 \]
Completing the square:
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 = -1 \]
\[ (x - 1)^2 + (y - 2)^2 = 6 \]
This shows that the center \(C_1\) is \((1, 2)\) and the radius \(r_1 = \sqrt{6}\).

2. **Circle 2**:
\[ x^2 + y^2 + 4y + 4x - 1 = 0 \]
Rearranging gives:
\[ (x^2 + 4x) + (y^2 + 4y) = 1 \]
Completing the square:
\[ (x + 2)^2 - 4 + (y + 2)^2 - 4 = 1 \]
\[ (x + 2)^2 + (y + 2)^2 = 9 \]
This shows that the center \(C_2\) is \((-2, -2)\) and the radius \(r_2 = 3\).

### Step 2: Find the distance between the centers of the circles

Using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of the centers \(C_1(1, 2)\) and \(C_2(-2, -2)\):
\[ d = \sqrt{((-2) - 1)^2 + ((-2) - 2)^2} \]
Calculating:
\[ d = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 3: Determine the relationship between the circles

Now we compare the distance \(d\) with the sum of the radii:
\[ r_1 + r_2 = \sqrt{6} + 3 \]
Since \(\sqrt{6} \approx 2.45\), we have:
\[ r_1 + r_2 \approx 2.45 + 3 = 5.45 \]
Since \(d = 5\) and \(r_1 + r_2 \approx 5.45\), we find that:
\[ d < r_1 + r_2 \]
This indicates that the circles touch externally.

### Step 4: Find the equation of the common tangent

The equation of the common tangent can be found using the formula:
\[ s_1 - s_2 = 0 \]
Where \(s_1\) and \(s_2\) are the equations of the circles.

Substituting the equations:
\[ (x^2 + y^2 - 2x - 4y + 1) - (x^2 + y^2 + 4x + 4y - 1) = 0 \]
This simplifies to:
\[ -6x - 8y + 2 = 0 \]
Dividing through by -2 gives:
\[ 3x + 4y - 1 = 0 \]

### Final Result

Thus, the equation of the common tangent is:
\[ 3x + 4y - 1 = 0 \]