If the circles `x^(2)+y^(2)=a^(2), x^(2)+y^(2)-6x-8y+9=0` touch externally then a=

Options

A. 4
B. 3
C. 2
D. 1

Answer

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

### Step-by-Step Solution:

1. **Identify the first circle:**
The equation of the first circle is given by:
\[ x^2 + y^2 = a^2 \]
From this, we can determine:
- Center \( C_1 = (0, 0) \)
- Radius \( r_1 = a \)

2. **Rewrite the second circle's equation:**
The second circle is given by:
\[ x^2 + y^2 - 6x - 8y + 9 = 0 \]
We can rewrite this in standard form by completing the square:
\[ (x^2 - 6x) + (y^2 - 8y) + 9 = 0 \]
Completing the square for \( x \):
\[ x^2 - 6x = (x - 3)^2 - 9 \]
Completing the square for \( y \):
\[ y^2 - 8y = (y - 4)^2 - 16 \]
Substituting these back into the equation gives:
\[ (x - 3)^2 - 9 + (y - 4)^2 - 16 + 9 = 0 \]
Simplifying this, we have:
\[ (x - 3)^2 + (y - 4)^2 - 16 = 0 \]
Thus, the equation becomes:
\[ (x - 3)^2 + (y - 4)^2 = 16 \]
From this, we can determine:
- Center \( C_2 = (3, 4) \)
- Radius \( r_2 = \sqrt{16} = 4 \)

3. **Calculate the distance between the centers:**
The distance \( d \) between the centers \( C_1 \) and \( C_2 \) is given by:
\[ d = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

4. **Use the condition for external tangency:**
For two circles to touch externally, the following condition must hold:
\[ d = r_1 + r_2 \]
Substituting the values we have:
\[ 5 = a + 4 \]

5. **Solve for \( a \):**
Rearranging the equation gives:
\[ a = 5 - 4 = 1 \]

Thus, the value of \( a \) is \( 1 \).

### Final Answer:
\[ a = 1 \]