If the circles `x^(2) + y^(2) = k` and `x^(2) + y^(2) + 8x - 6y + 9 = 0 ` touch externally, then the value of `k` is

Options

A. 1
B. `-1`
C. 9
D. 81

Answer

To solve the problem, we need to find the value of \( k \) such that the circles defined by the equations \( x^2 + y^2 = k \) and \( x^2 + y^2 + 8x - 6y + 9 = 0 \) touch each other externally.

### Step-by-Step Solution:

1. **Identify the first circle:**
The first circle is given by the equation:
\[ x^2 + y^2 = k \]
- Center: \( (0, 0) \)
- Radius: \( r_1 = \sqrt{k} \)

2. **Identify the second circle:**
The second circle is given by:
\[ x^2 + y^2 + 8x - 6y + 9 = 0 \]
We can rewrite this in standard form by completing the square.

- Rearranging gives:
\[ x^2 + 8x + y^2 - 6y + 9 = 0 \]

- Completing the square for \( x \):
\[ x^2 + 8x = (x + 4)^2 - 16 \]

- Completing the square for \( y \):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

- Substituting back, we have:
\[ (x + 4)^2 - 16 + (y - 3)^2 - 9 + 9 = 0 \]
Simplifying:
\[ (x + 4)^2 + (y - 3)^2 - 16 = 0 \implies (x + 4)^2 + (y - 3)^2 = 16 \]

- Center: \( (-4, 3) \)
- Radius: \( r_2 = \sqrt{16} = 4 \)

3. **Condition for external tangency:**
For two circles to touch externally, the distance between their centers must equal the sum of their radii:
\[ d = r_1 + r_2 \]

4. **Calculate the distance between the centers:**
The distance \( d \) between the centers \( (0, 0) \) and \( (-4, 3) \) is given by:
\[ d = \sqrt{(-4 - 0)^2 + (3 - 0)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

5. **Set up the equation:**
According to the tangency condition:
\[ d = r_1 + r_2 \implies 5 = \sqrt{k} + 4 \]

6. **Solve for \( k \):**
Rearranging gives:
\[ \sqrt{k} = 5 - 4 = 1 \]
Squaring both sides:
\[ k = 1^2 = 1 \]

### Final Answer:
The value of \( k \) is \( \boxed{1} \).