The circle `x^(2)+y^(2) - 2x+c=0` touches y-axis, then c =

Options

A. 1
B. 0
C. `-1`
D. none

Answer

To solve the problem, we need to determine the value of \( c \) such that the circle given by the equation \( x^2 + y^2 - 2x + c = 0 \) touches the y-axis.

### Step-by-Step Solution:

1. **Identify the Circle's Center and Radius**:
The general form of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + h = 0 \]
From the given equation \( x^2 + y^2 - 2x + c = 0 \), we can identify:
- \( 2g = -2 \) → \( g = -1 \)
- \( 2f = 0 \) → \( f = 0 \)
- \( h = c \)

Thus, the center of the circle is \( (-g, -f) = (1, 0) \).

2. **Calculate the Radius**:
The radius \( r \) of the circle can be calculated using the formula:
\[ r = \sqrt{g^2 + f^2 - h} \]
Substituting the values we have:
\[ r = \sqrt{(-1)^2 + 0^2 - c} = \sqrt{1 - c} \]

3. **Condition for the Circle to Touch the Y-axis**:
For the circle to touch the y-axis, the distance from the center of the circle to the y-axis must equal the radius. The distance from the center \( (1, 0) \) to the y-axis (which is at \( x = 0 \)) is \( 1 \).

Therefore, we set up the equation:
\[ r = 1 \]
Substituting the expression for the radius:
\[ \sqrt{1 - c} = 1 \]

4. **Solve for \( c \)**:
Squaring both sides of the equation gives:
\[ 1 - c = 1 \]
Rearranging this, we find:
\[ c = 0 \]

### Conclusion:
Thus, the value of \( c \) such that the circle touches the y-axis is:
\[ \boxed{0} \]