The circle `4x^(2)+4y^(2)-12x-12y+9=0`

Options

A. touches both the axes
B. touches the x-axis only
C. touches the y-axis only
D. does not touche the axes

Answer

To solve the problem, we need to analyze the given equation of the circle and determine its center and radius. The equation provided is:

\[ 4x^2 + 4y^2 - 12x - 12y + 9 = 0 \]

### Step 1: Simplify the equation
First, we divide the entire equation by 4 to simplify it:

\[ x^2 + y^2 - 3x - 3y + \frac{9}{4} = 0 \]

### Step 2: Rearranging the equation
Next, we rearrange the equation to group the \(x\) and \(y\) terms:

\[ x^2 - 3x + y^2 - 3y = -\frac{9}{4} \]

### Step 3: Completing the square for \(x\) and \(y\)
Now we complete the square for both \(x\) and \(y\):

For \(x^2 - 3x\):
- Take half of the coefficient of \(x\) (which is \(-3\)), square it: \(\left(-\frac{3}{2}\right)^2 = \frac{9}{4}\).
- Thus, we can write:
\[ x^2 - 3x = \left(x - \frac{3}{2}\right)^2 - \frac{9}{4} \]

For \(y^2 - 3y\):
- Similarly, we do the same for \(y\):
\[ y^2 - 3y = \left(y - \frac{3}{2}\right)^2 - \frac{9}{4} \]

### Step 4: Substitute back into the equation
Substituting these completed squares back into the equation gives us:

\[ \left(x - \frac{3}{2}\right)^2 - \frac{9}{4} + \left(y - \frac{3}{2}\right)^2 - \frac{9}{4} = -\frac{9}{4} \]

### Step 5: Simplifying the equation
Now, we simplify the equation:

\[ \left(x - \frac{3}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 - \frac{9}{2} = -\frac{9}{4} \]

Adding \(\frac{9}{2}\) to both sides:

\[ \left(x - \frac{3}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 = \frac{9}{4} \]

### Step 6: Identify the center and radius
From the standard form of the circle \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center \((h, k) = \left(\frac{3}{2}, \frac{3}{2}\right)\)
- Radius \(r = \sqrt{\frac{9}{4}} = \frac{3}{2}\)

### Step 7: Determine the position relative to the axes
Since the center is \(\left(\frac{3}{2}, \frac{3}{2}\right)\) and the radius is \(\frac{3}{2}\), we can analyze how the circle interacts with the axes:
- The circle touches the x-axis at \(y = 0\) (since the center is at \(y = \frac{3}{2}\) and the radius is also \(\frac{3}{2}\)).
- The circle touches the y-axis at \(x = 0\) (since the center is at \(x = \frac{3}{2}\) and the radius is also \(\frac{3}{2}\)).

### Conclusion
Thus, the circle touches both axes.