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

Options

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

Answer

To find the value of \( a \) such that the circle given by the equation

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

touches the y-axis, we will follow these steps:

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

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

where \( (h, k) \) is the center and \( r \) is the radius. We will complete the square for the \( x \) and \( y \) terms.

### Step 2: Complete the Square for \( x \) and \( y \)
The given equation is:

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

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:

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

Simplifying this:

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

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

This simplifies to:

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

### Step 3: Identify the Center and Radius
From the standard form, we can identify the center \( (h, k) \) and the radius \( r \):

- Center \( (h, k) = (3, 4) \)
- Radius \( r = a \)

### Step 4: Condition for the Circle to Touch the Y-axis
For the circle to touch the y-axis, the distance from the center to the y-axis must equal the radius. The distance from the center \( (3, 4) \) to the y-axis (which is at \( x = 0 \)) is \( |h| = 3 \).

Thus, we set up the equation:

\[ r = |h| \implies a = 3 \]

### Final Answer
Thus, the value of \( a \) is:

\[ \boxed{3} \]