If the circle `x ^(2) +y ^(2) - 6x - 8y + (25 -a ^(2)) =0` touches the axis of x, then a equals

Options

A. 0
B. `pm4`
C. `pm2`
D. `pm3`

Answer

To solve the problem, we need to determine the value of \( a \) such that the circle defined by the equation

\[ x^2 + y^2 - 6x - 8y + (25 - a^2) = 0 \]

touches the x-axis.

### Step 1: Rewrite the Circle Equation
First, we rewrite the circle equation in standard form. The general form of a circle is given by:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

where \( (h, k) \) is the center and \( r \) is the radius. We can rearrange the given equation:

\[ x^2 - 6x + y^2 - 8y + (25 - a^2) = 0 \]

### Step 2: Complete the Square
Next, we complete the square for the \( x \) and \( y \) terms.

For \( x^2 - 6x \):
\[ x^2 - 6x = (x - 3)^2 - 9 \]

For \( y^2 - 8y \):
\[ y^2 - 8y = (y - 4)^2 - 16 \]

Substituting these back into the equation gives:

\[ (x - 3)^2 - 9 + (y - 4)^2 - 16 + (25 - a^2) = 0 \]

### Step 3: Simplify the Equation
Now, simplify the equation:

\[ (x - 3)^2 + (y - 4)^2 - 9 - 16 + 25 - a^2 = 0 \]
\[ (x - 3)^2 + (y - 4)^2 + (0 - a^2) = 0 \]

This simplifies to:

\[ (x - 3)^2 + (y - 4)^2 = a^2 \]

### Step 4: Identify the Center and Radius
From the standard form, we can identify the center of the circle as \( (3, 4) \) and the radius as \( r = a \).

### Step 5: Condition for Touching the x-axis
For the circle to touch the x-axis, the distance from the center to the x-axis must equal the radius. The distance from the center \( (3, 4) \) to the x-axis is the y-coordinate of the center, which is \( 4 \).

Thus, we set the radius equal to this distance:

\[ a = 4 \]

### Step 6: Conclusion
Therefore, the value of \( a \) is:

\[ \boxed{4} \]