The circles `x^(2)+y^(2)+2x-4y+4=0` and `x^(2)+y^(2)-2x-4y+4=0` are such that they

Options

A. touch internally
B. touch externally
C. intersect on axis of y
D. touch at (0, 2)

Answer

To solve the problem, we need to analyze the two given circles and determine their relationship based on their centers and radii.

### Step 1: Write the equations of the circles in standard form
The given equations of the circles are:
1. \( x^2 + y^2 + 2x - 4y + 4 = 0 \)
2. \( x^2 + y^2 - 2x - 4y + 4 = 0 \)

### Step 2: Identify the coefficients and find the centers and radii
For the first circle:
- The general form of a circle is \( x^2 + y^2 + 2gx + 2fy + c = 0 \).
- Here, \( g_1 = 1 \), \( f_1 = -2 \), and \( c_1 = 4 \).
- The center \( C_1 \) is given by \( (-g_1, -f_1) = (-1, 2) \).
- The radius \( r_1 \) is calculated as \( r_1 = \sqrt{g_1^2 + f_1^2 - c_1} = \sqrt{1^2 + (-2)^2 - 4} = \sqrt{1 + 4 - 4} = \sqrt{1} = 1 \).

For the second circle:
- Here, \( g_2 = -1 \), \( f_2 = -2 \), and \( c_2 = 4 \).
- The center \( C_2 \) is given by \( (-g_2, -f_2) = (1, 2) \).
- The radius \( r_2 \) is calculated as \( r_2 = \sqrt{g_2^2 + f_2^2 - c_2} = \sqrt{(-1)^2 + (-2)^2 - 4} = \sqrt{1 + 4 - 4} = \sqrt{1} = 1 \).

### Step 3: Calculate the distance between the centers
The distance \( d \) between the centers \( C_1(-1, 2) \) and \( C_2(1, 2) \) is calculated as:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(1 - (-1))^2 + (2 - 2)^2} = \sqrt{(1 + 1)^2 + 0} = \sqrt{4} = 2. \]

### Step 4: Compare the distance with the sum of the radii
The sum of the radii \( r_1 + r_2 = 1 + 1 = 2 \).

### Step 5: Determine the relationship between the circles
Since the distance between the centers \( d = 2 \) is equal to the sum of the radii \( r_1 + r_2 = 2 \), the circles touch externally.

### Conclusion
The circles touch externally.