Find the direct common tangents of the circles `x^(2) + y^(2) +22x -4y -100 =0 and x^(2) + y^(2) -22x + 4y +100 =0 `

Answer

To find the direct common tangents of the given circles, we will follow these steps:

### Step 1: Write the equations of the circles in standard form
The given equations of the circles are:
1. \( x^2 + y^2 + 22x - 4y - 100 = 0 \)
2. \( x^2 + y^2 - 22x + 4y + 100 = 0 \)

### Step 2: Identify the center and radius of each circle
For a circle in the form \( x^2 + y^2 + 2gx + 2fy + c = 0 \):
- Center \( C(-g, -f) \)
- Radius \( r = \sqrt{g^2 + f^2 - c} \)

**Circle 1:**
- Coefficients: \( g = 11, f = -2, c = -100 \)
- Center \( C_1 = (-11, 2) \)
- Radius \( r_1 = \sqrt{11^2 + (-2)^2 - (-100)} = \sqrt{121 + 4 + 100} = \sqrt{225} = 15 \)

**Circle 2:**
- Coefficients: \( g = -11, f = 2, c = 100 \)
- Center \( C_2 = (11, -2) \)
- Radius \( r_2 = \sqrt{(-11)^2 + 2^2 - 100} = \sqrt{121 + 4 - 100} = \sqrt{25} = 5 \)

### Step 3: Calculate the distance between the centers of the circles
The distance \( d \) between the centers \( C_1 \) and \( C_2 \) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(11 - (-11))^2 + (-2 - 2)^2} = \sqrt{(22)^2 + (-4)^2} = \sqrt{484 + 16} = \sqrt{500} = 10\sqrt{5} \]

### Step 4: Check the condition for direct common tangents
For direct common tangents to exist, the following condition must hold:
\[ d > r_1 + r_2 \]
Calculating \( r_1 + r_2 \):
\[ r_1 + r_2 = 15 + 5 = 20 \]
Since \( 10\sqrt{5} \approx 22.36 > 20 \), direct common tangents exist.

### Step 5: Use the formula for the slopes of the tangents
Let the slopes of the tangents be \( m_1 \) and \( m_2 \). The slopes can be found using the formula:
\[ \frac{r_1 - r_2}{d} \text{ and } \frac{r_1 + r_2}{d} \]
Calculating:
\[ \text{Slope 1: } m_1 = \frac{15 - 5}{10\sqrt{5}} = \frac{10}{10\sqrt{5}} = \frac{1}{\sqrt{5}} \]
\[ \text{Slope 2: } m_2 = \frac{15 + 5}{10\sqrt{5}} = \frac{20}{10\sqrt{5}} = \frac{2}{\sqrt{5}} \]

### Step 6: Write the equations of the tangents
Using the point-slope form of a line \( y - y_1 = m(x - x_1) \) where \( (x_1, y_1) \) is a point on the tangent line (we can use the point \( P(22, -4) \)):
1. For slope \( m_1 = \frac{1}{\sqrt{5}} \):
\[ y + 4 = \frac{1}{\sqrt{5}}(x - 22) \]
Rearranging gives:
\[ \sqrt{5}y + 4\sqrt{5} = x - 22 \Rightarrow x - \sqrt{5}y - 22 - 4\sqrt{5} = 0 \]

2. For slope \( m_2 = \frac{2}{\sqrt{5}} \):
\[ y + 4 = \frac{2}{\sqrt{5}}(x - 22) \]
Rearranging gives:
\[ \sqrt{5}y + 4\sqrt{5} = 2x - 44 \Rightarrow 2x - \sqrt{5}y - 44 - 4\sqrt{5} = 0 \]

### Final Answer
The equations of the direct common tangents are:
1. \( x - \sqrt{5}y - (22 + 4\sqrt{5}) = 0 \)
2. \( 2x - \sqrt{5}y - (44 + 4\sqrt{5}) = 0 \)