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

Options

A. 1
B. 2
C. 3
D. 6

Answer

To solve the problem of determining the value of \( a \) for which the circles \( x^2 + y^2 = a \) and \( x^2 + y^2 - 6x - 8y + 9 = 0 \) touch externally, we can follow these steps:

### Step 1: Identify the centers and radii of the circles.

1. **Circle 1**: The equation \( x^2 + y^2 = a \) represents a circle centered at \( (0, 0) \) with radius \( r_1 = \sqrt{a} \).

2. **Circle 2**: The equation \( x^2 + y^2 - 6x - 8y + 9 = 0 \) can be rewritten by completing the square:
\[ (x^2 - 6x) + (y^2 - 8y) + 9 = 0 \]
Completing the square for \( x \) and \( y \):
\[ (x - 3)^2 - 9 + (y - 4)^2 - 16 + 9 = 0 \]
This simplifies to:
\[ (x - 3)^2 + (y - 4)^2 = 16 \]
Thus, Circle 2 is centered at \( (3, 4) \) with radius \( r_2 = 4 \).

### Step 2: Calculate the distance between the centers.

The distance \( d \) between the centers of the two circles can be calculated using the distance formula:
\[ d = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 3: Set up the equation for external tangency.

For the circles to touch externally, the distance between their centers must equal the sum of their radii:
\[ d = r_1 + r_2 \]
Substituting the known values:
\[ 5 = \sqrt{a} + 4 \]

### Step 4: Solve for \( a \).

Rearranging the equation gives:
\[ \sqrt{a} = 5 - 4 = 1 \]
Squaring both sides results in:
\[ a = 1^2 = 1 \]

### Conclusion

Thus, the value of \( a \) for which the circles touch externally is:
\[ \boxed{1} \]