The circle `x^2+y^2-4x-4y+4=0` is

Options

A. touches x axis only
B. touches both axis
C. passes through the origin
D. touches y axis only

Answer

To solve the problem, we need to analyze the given equation of the circle and determine its properties regarding its interaction with the axes.

### Step-by-Step Solution:

1. **Rewrite the Circle Equation:**
The given equation of the circle is:
\[ x^2 + y^2 - 4x - 4y + 4 = 0 \]
We can rearrange this to the standard form of a circle.

2. **Complete the Square:**
We will complete the square for both \(x\) and \(y\).

- For \(x\):
\[ x^2 - 4x \rightarrow (x - 2)^2 - 4 \]

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

Putting these into the equation gives:
\[ (x - 2)^2 - 4 + (y - 2)^2 - 4 + 4 = 0 \]
Simplifying this, we have:
\[ (x - 2)^2 + (y - 2)^2 - 4 = 0 \]
or
\[ (x - 2)^2 + (y - 2)^2 = 4 \]

3. **Identify the Center and Radius:**
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center \((h, k) = (2, 2)\)
- Radius \(r = \sqrt{4} = 2\)

4. **Check the Circle's Interaction with the Axes:**
- **Touching the X-axis:** A circle touches the x-axis if the distance from the center to the x-axis is equal to the radius. The distance from the center \((2, 2)\) to the x-axis is \(2\), which is equal to the radius. Thus, it touches the x-axis.
- **Touching the Y-axis:** Similarly, the distance from the center \((2, 2)\) to the y-axis is also \(2\), which is equal to the radius. Thus, it touches the y-axis.

5. **Conclusion:**
Since the circle touches both the x-axis and the y-axis, the correct option is:
- **Touches both axes.**

### Final Answer:
The circle touches both axes.