The chords through (2,1) to the circle `x^2-2x+y^2-2y+1=0` are bisected at the point `(alpha,1/2)` then value of `(2alpha)/3` is ___

Answer

To solve the problem, we need to find the value of \( \frac{2\alpha}{3} \) given that the chords through the point \( (2, 1) \) to the circle defined by the equation \( x^2 - 2x + y^2 - 2y + 1 = 0 \) are bisected at the point \( (\alpha, \frac{1}{2}) \).

### Step-by-Step Solution:

1. **Rewrite the Circle Equation**:
The given circle equation is:
\[ x^2 - 2x + y^2 - 2y + 1 = 0 \]
We can complete the square for both \( x \) and \( y \):
\[ (x-1)^2 + (y-1)^2 = 1 \]
This represents a circle with center \( (1, 1) \) and radius \( 1 \).

2. **Equation of the Chord**:
The equation of the chord that passes through the point \( (2, 1) \) and is bisected at \( (\alpha, \frac{1}{2}) \) can be derived using the formula for the chord of a circle:
\[ T = S_1 \]
where \( T \) is the equation of the chord and \( S_1 \) is the equation of the circle evaluated at the midpoint \( (\alpha, \frac{1}{2}) \).

The equation of the chord can be expressed as:
\[ \alpha(x - 2) + \frac{1}{2}(y - 1) = 0 \]
Rearranging gives:
\[ \alpha x + \frac{1}{2}y - \alpha \cdot 2 - \frac{1}{2} = 0 \]

3. **Substituting the Midpoint into the Circle Equation**:
Substitute \( (2, 1) \) into the equation of the circle:
\[ (2-1)^2 + (1-1)^2 = 1 \]
This confirms that \( (2, 1) \) lies on the circle.

4. **Using the Chord Condition**:
Since the chord passes through \( (2, 1) \), we substitute \( x = 2 \) and \( y = 1 \) into the chord equation:
\[ \alpha(2 - 2) + \frac{1}{2}(1 - 1) = 0 \]
This does not provide new information, so we proceed to find the intersection points of the chord with the circle.

5. **Finding the Intersection Points**:
The intersection of the chord with the circle gives us the points where the chord intersects the circle. We set up the equation:
\[ (x-1)^2 + (y-1)^2 = 1 \]
and substitute \( y = \frac{1}{2} \) into the circle equation:
\[ (x-1)^2 + \left(\frac{1}{2} - 1\right)^2 = 1 \]
Simplifying gives:
\[ (x-1)^2 + \left(-\frac{1}{2}\right)^2 = 1 \]
\[ (x-1)^2 + \frac{1}{4} = 1 \]
\[ (x-1)^2 = \frac{3}{4} \]
Taking square roots:
\[ x - 1 = \pm \frac{\sqrt{3}}{2} \]
Thus, we find:
\[ x = 1 \pm \frac{\sqrt{3}}{2} \]

6. **Finding the Value of \( \alpha \)**:
The midpoint \( \alpha \) must be the average of the x-coordinates of the intersection points:
\[ \alpha = \frac{(1 + \frac{\sqrt{3}}{2}) + (1 - \frac{\sqrt{3}}{2})}{2} = \frac{2}{2} = 1 \]

7. **Calculating \( \frac{2\alpha}{3} \)**:
Now we substitute \( \alpha = 1 \) into \( \frac{2\alpha}{3} \):
\[ \frac{2 \cdot 1}{3} = \frac{2}{3} \]

### Final Answer:
The value of \( \frac{2\alpha}{3} \) is \( \frac{2}{3} \).