The intercept on the line y=x by the circle `x^(2)+y^(2)-2x=0` is AB. Equation of the circle on AB as diameter is:

Options

A. `x^(2)+y^(2)-x-y=0`
B. `x^(2)+y^(2)-x+y=0`
C. `x^(2)+y^(2)+x+y=0`
D. `x^(2)+y^(2)+x-y=0`

Answer

To solve the problem, we need to find the equation of the circle that has the line segment AB (the intercept on the line y = x by the given circle) as its diameter.

### Step-by-Step Solution:

1. **Identify the given circle's equation**:
The equation of the circle is given as:
\[ x^2 + y^2 - 2x = 0 \]

2. **Rearrange the circle's equation**:
We can rewrite the equation as:
\[ x^2 - 2x + y^2 = 0 \]
To make it easier to analyze, we can complete the square for the x-term:
\[ (x - 1)^2 + y^2 = 1 \]
This shows that the circle is centered at (1, 0) with a radius of 1.

3. **Find the points of intersection with the line y = x**:
Substitute \(y = x\) into the circle's equation:
\[ (x - 1)^2 + x^2 = 1 \]
Expanding this gives:
\[ (x^2 - 2x + 1) + x^2 = 1 \]
Simplifying:
\[ 2x^2 - 2x + 1 - 1 = 0 \implies 2x^2 - 2x = 0 \]
Factoring out \(2x\):
\[ 2x(x - 1) = 0 \]
Thus, \(x = 0\) or \(x = 1\). Therefore, the points of intersection are:
\[ (0, 0) \quad \text{and} \quad (1, 1) \]

4. **Determine the endpoints of the diameter AB**:
The points of intersection \(A(0, 0)\) and \(B(1, 1)\) serve as the endpoints of the diameter of the new circle.

5. **Find the center and radius of the new circle**:
The center of the circle with AB as the diameter is the midpoint of A and B:
\[ \text{Midpoint} = \left(\frac{0 + 1}{2}, \frac{0 + 1}{2}\right) = \left(\frac{1}{2}, \frac{1}{2}\right) \]
The radius is half the distance between A and B:
\[ \text{Distance AB} = \sqrt{(1 - 0)^2 + (1 - 0)^2} = \sqrt{2} \]
Therefore, the radius is:
\[ \text{Radius} = \frac{\sqrt{2}}{2} \]

6. **Write the equation of the new circle**:
The standard form of the circle's equation is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = \frac{1}{2}\), \(k = \frac{1}{2}\), and \(r = \frac{\sqrt{2}}{2}\):
\[ \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{\sqrt{2}}{2}\right)^2 \]
Simplifying gives:
\[ \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{1}{2} \]

7. **Expand the equation**:
Expanding the left side:
\[ \left(x^2 - x + \frac{1}{4}\right) + \left(y^2 - y + \frac{1}{4}\right) = \frac{1}{2} \]
Combining terms:
\[ x^2 + y^2 - x - y + \frac{1}{2} = \frac{1}{2} \]
Thus, the final equation simplifies to:
\[ x^2 + y^2 - x - y = 0 \]

### Final Answer:
The equation of the circle on AB as diameter is:
\[ x^2 + y^2 - x - y = 0 \]