If the area of the circle `4x^(2)+4y^(2)-8x+16y+k=0` is `9pi` square units, then the value of k is

Options

A. 4
B. 16
C. -16
D. `+-16`

Explanation

To solve the problem, we need to find the value of \( k \) in the equation of the circle given by:

\[ 4x^2 + 4y^2 - 8x + 16y + k = 0 \]

We know that the area of the circle is \( 9\pi \) square units.

### Step 1: Rewrite the equation in standard form
First, we divide the entire equation by 4 to simplify it:

\[ x^2 + y^2 - 2x + 4y + \frac{k}{4} = 0 \]

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

For \( x \):
\[ x^2 - 2x = (x - 1)^2 - 1 \]

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

Substituting these back into the equation gives:

\[ (x - 1)^2 - 1 + (y + 2)^2 - 4 + \frac{k}{4} = 0 \]

Combining the constants:

\[ (x - 1)^2 + (y + 2)^2 + \frac{k}{4} - 5 = 0 \]

### Step 3: Rearranging the equation
Rearranging the equation, we have:

\[ (x - 1)^2 + (y + 2)^2 = 5 - \frac{k}{4} \]

### Step 4: Identify the radius
From the standard form of the circle equation \( (x - h)^2 + (y - k)^2 = r^2 \), we can identify:

\[ r^2 = 5 - \frac{k}{4} \]

### Step 5: Use the area of the circle
The area \( A \) of a circle is given by:

\[ A = \pi r^2 \]

Given that the area is \( 9\pi \):

\[ \pi r^2 = 9\pi \]

Dividing both sides by \( \pi \):

\[ r^2 = 9 \]

### Step 6: Set the equations equal
Now we can set the two expressions for \( r^2 \) equal to each other:

\[ 5 - \frac{k}{4} = 9 \]

### Step 7: Solve for \( k \)
Rearranging gives:

\[ -\frac{k}{4} = 9 - 5 \]

\[ -\frac{k}{4} = 4 \]

Multiplying both sides by -4:

\[ k = -16 \]

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

\[ \boxed{-16} \]