What is the equation of circle which touches the lines x = 0 ?

Options

A. `x^(2) + y^(2) + 2x + 2y +1= 0`
B. `x^(2) + y^(2) - 4x - 4y + 1 = 0`
C. `x^(2) + y^(2) -2x - 2y + 1 = 0`
D. None of these

Answer

To find the equation of a circle that touches the line \( x = 0 \), we need to understand that this line represents the y-axis. A circle that touches the y-axis will have its center at a distance equal to its radius from the y-axis.

### Step-by-Step Solution:

1. **Identify the Center and Radius**:
Let the center of the circle be at point \( (h, k) \) and the radius be \( r \). Since the circle touches the line \( x = 0 \), the distance from the center to this line must equal the radius. Thus, we have:
\[ |h| = r \]

2. **General Equation of a Circle**:
The general equation of a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]

3. **Substituting the Condition**:
From the condition \( |h| = r \), we can express \( r \) in terms of \( h \):
\[ r = |h| \]

4. **Substituting into the Circle Equation**:
Substitute \( r \) into the equation of the circle:
\[ (x - h)^2 + (y - k)^2 = h^2 \]

5. **Final Form of the Equation**:
The equation of the circle that touches the line \( x = 0 \) can be written as:
\[ (x - h)^2 + (y - k)^2 = h^2 \]
where \( h \) can be any real number, and \( k \) can also be any real number.

### Conclusion:
Thus, the equation of the circle that touches the line \( x = 0 \) is:
\[ (x - h)^2 + (y - k)^2 = h^2 \]
where \( h \) is the x-coordinate of the center and \( k \) is the y-coordinate of the center.