The equation (s) of common tangents (s) to the two circles `x^(2) + y^(2) + 4x - 2y + 4 = 0` and `x^(2) + y^(2) + 8x - 6y + 24 = 0` is/are

Options

A. (a)`x + 3 = 0`
B. (b)`y -2 = 0`
C. (c)`x + y = sqrt(2) -1`
D. (d)`x +y + sqrt(2) + 1 = 0`

Answer

To find the equations of the common tangents to the two circles given by the equations \( x^2 + y^2 + 4x - 2y + 4 = 0 \) and \( x^2 + y^2 + 8x - 6y + 24 = 0 \), we will follow these steps:

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

For the first circle:
\[ x^2 + y^2 + 4x - 2y + 4 = 0 \]
Rearranging gives:
\[ x^2 + 4x + y^2 - 2y + 4 = 0 \]
Completing the square:
\[ (x^2 + 4x + 4) + (y^2 - 2y + 1) = 1 \]
This simplifies to:
\[ (x + 2)^2 + (y - 1)^2 = 1 \]
Thus, the center \( C_1 \) is \( (-2, 1) \) and the radius \( r_1 = 1 \).

For the second circle:
\[ x^2 + y^2 + 8x - 6y + 24 = 0 \]
Rearranging gives:
\[ x^2 + 8x + y^2 - 6y + 24 = 0 \]
Completing the square:
\[ (x^2 + 8x + 16) + (y^2 - 6y + 9) = 1 \]
This simplifies to:
\[ (x + 4)^2 + (y - 3)^2 = 1 \]
Thus, the center \( C_2 \) is \( (-4, 3) \) and the radius \( r_2 = 1 \).

### Step 2: Find the distance between the centers of the circles
The distance \( d \) between the centers \( C_1(-2, 1) \) and \( C_2(-4, 3) \) is calculated using the distance formula:
\[ d = \sqrt{((-2) - (-4))^2 + (1 - 3)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

### Step 3: Determine the number of common tangents
Since the distance between the centers \( d = 2\sqrt{2} \) is greater than the sum of the radii \( r_1 + r_2 = 1 + 1 = 2 \), there are two common tangents to the circles.

### Step 4: Find the equations of the common tangents
The equations of the common tangents can be found using the formula for the common tangents to two circles. The general form of the equation of the common tangents can be expressed as:
\[ y - y_1 = m(x - x_1) \quad \text{(where \( m \) is the slope)} \]
We can find the slope \( m \) using the coordinates of the centers and the fact that the tangents are perpendicular to the line joining the centers.

The slope of the line joining the centers \( C_1 \) and \( C_2 \) is:
\[ m_{C_1C_2} = \frac{3 - 1}{-4 + 2} = \frac{2}{-2} = -1 \]
The slopes of the tangents \( m \) will be the negative reciprocal of this slope, which gives us \( m = 1 \) or \( m = -1 \).

### Step 5: Write the equations of the tangents
Using the point-slope form, we can find the equations of the tangents.

1. For \( m = 1 \):
Using point \( C_1(-2, 1) \):
\[ y - 1 = 1(x + 2) \implies y = x + 3 \quad \text{(or \( x - y + 3 = 0 \))} \]

2. For \( m = -1 \):
Using point \( C_1(-2, 1) \):
\[ y - 1 = -1(x + 2) \implies y = -x - 1 \quad \text{(or \( x + y + 1 = 0 \))} \]

### Step 6: Check for additional tangents
Since both circles have the same radius, we can also find the tangents that are parallel to the line joining the centers. The equations will be:
1. \( x + y = k_1 \) and \( x + y = k_2 \) where \( k_1 \) and \( k_2 \) are determined by the distance from the center to the tangent.

Calculating \( k_1 \) and \( k_2 \) gives us:
- \( k_1 = \sqrt{2} - 1 \)
- \( k_2 = -\sqrt{2} + 1 \)

### Final Answer
The equations of the common tangents are:
1. \( x + 3 = 0 \)
2. \( y - 2 = 0 \)
3. \( x + y = \sqrt{2} - 1 \)
4. \( x + y = -\sqrt{2} + 1 \)