If circles ` x^(2) + y^(2) = 9 " and " x^(2) + y^(2) + 2ax + 2y + 1 = 0 `
touch each other , then a =

Options

A. 0
B. 1
C. `pm (4)/(3)`
D. `pm (3)/(4)`

Answer

To solve the problem, we need to find the value of \( a \) such that the circles \( x^2 + y^2 = 9 \) and \( x^2 + y^2 + 2ax + 2y + 1 = 0 \) touch each other.

### Step-by-Step Solution:

1. **Identify the first circle:**
The first circle is given by the equation:
\[ x^2 + y^2 = 9 \]
This can be rewritten in the standard form:
\[ x^2 + y^2 + 0x + 0y - 9 = 0 \]
From this, we can identify:
- Center \( C_1 = (0, 0) \)
- Radius \( R_1 = \sqrt{0^2 + 0^2 + 9} = 3 \)

2. **Identify the second circle:**
The second circle is given by:
\[ x^2 + y^2 + 2ax + 2y + 1 = 0 \]
This can be rewritten in the standard form:
\[ x^2 + y^2 + 2ax + 2y + 1 = 0 \]
From this, we can identify:
- Center \( C_2 = (-a, -1) \)
- Radius \( R_2 = \sqrt{(a)^2 + (1)^2 - 1} = \sqrt{a^2} = |a| \)

3. **Calculate the distance between the centers:**
The distance \( d \) between the centers \( C_1 \) and \( C_2 \) is given by:
\[ d = \sqrt{(-a - 0)^2 + (-1 - 0)^2} = \sqrt{a^2 + 1} \]

4. **Set up the equation for external tangency:**
For the circles to touch externally, the following condition must hold:
\[ R_1 + R_2 = d \]
Substituting the values we found:
\[ 3 + |a| = \sqrt{a^2 + 1} \]

5. **Square both sides to eliminate the square root:**
\[ (3 + |a|)^2 = a^2 + 1 \]
Expanding the left side:
\[ 9 + 6|a| + a^2 = a^2 + 1 \]
Simplifying:
\[ 9 + 6|a| = 1 \]
\[ 6|a| = 1 - 9 \]
\[ 6|a| = -8 \]
Since \( |a| \) cannot be negative, this case does not yield a valid solution.

6. **Set up the equation for internal tangency:**
For the circles to touch internally, the following condition must hold:
\[ |R_1 - R_2| = d \]
Substituting the values:
\[ |3 - |a|| = \sqrt{a^2 + 1} \]

7. **Consider two cases for \( |3 - |a|| \):**
- **Case 1:** \( 3 - a = \sqrt{a^2 + 1} \)
- **Case 2:** \( a - 3 = \sqrt{a^2 + 1} \)

**For Case 1:**
\[ 3 - a = \sqrt{a^2 + 1} \]
Squaring both sides:
\[ (3 - a)^2 = a^2 + 1 \]
Expanding:
\[ 9 - 6a + a^2 = a^2 + 1 \]
Simplifying:
\[ 9 - 6a = 1 \]
\[ -6a = -8 \implies a = \frac{4}{3} \]

**For Case 2:**
\[ a - 3 = \sqrt{a^2 + 1} \]
Squaring both sides:
\[ (a - 3)^2 = a^2 + 1 \]
Expanding:
\[ a^2 - 6a + 9 = a^2 + 1 \]
Simplifying:
\[ -6a + 9 = 1 \]
\[ -6a = -8 \implies a = \frac{4}{3} \]

### Final Answer:
The value of \( a \) is:
\[ \boxed{-\frac{4}{3}} \]