The locus of the midpoint of the chord of the circle `x^(2)+y^(2)-2x-2y-2=0` which makes an angle of `120^(@)` at the centre is

Options

A. `x^(2)+y^(2)-2x-2y-1=0`
B. `x^(2)+y^(2)-2x-2y=0`
C. `x^(2)+y^(2)+2x-2y+1=0`
D. `x^(2)+y^(2)-2x-2y+1=0`

Answer

To solve the problem, we need to find the locus of the midpoint of a chord of the circle given by the equation \( x^2 + y^2 - 2x - 2y - 2 = 0 \) that makes an angle of \( 120^\circ \) at the center of the circle.

### Step-by-Step Solution:

1. **Identify the Circle's Center and Radius**:
- The equation of the circle can be rewritten by completing the square:
\[ x^2 - 2x + y^2 - 2y - 2 = 0 \]
Completing the square for \( x \) and \( y \):
\[ (x-1)^2 + (y-1)^2 = 4 \]
- This shows that the center of the circle is \( (1, 1) \) and the radius is \( 2 \).

2. **Define the Midpoint of the Chord**:
- Let the midpoint of the chord be \( M(h, k) \). The distance from the center \( C(1, 1) \) to the midpoint \( M(h, k) \) is given by the distance formula:
\[ MC = \sqrt{(h - 1)^2 + (k - 1)^2} \]

3. **Using the Angle Condition**:
- The angle subtended by the chord at the center is \( 120^\circ \). Therefore, the angle subtended at the midpoint \( M \) is half of \( 120^\circ \), which is \( 60^\circ \).
- Using the cosine of the angle, we can relate the lengths:
\[ \cos(60^\circ) = \frac{1}{2} = \frac{MC}{r} \]
- Here, \( r = 2 \) (the radius of the circle), so:
\[ \frac{MC}{2} = \frac{1}{2} \implies MC = 1 \]

4. **Setting Up the Equation**:
- From the distance formula, we have:
\[ \sqrt{(h - 1)^2 + (k - 1)^2} = 1 \]
- Squaring both sides gives:
\[ (h - 1)^2 + (k - 1)^2 = 1 \]

5. **Substituting Back**:
- Replacing \( h \) with \( x \) and \( k \) with \( y \), we get:
\[ (x - 1)^2 + (y - 1)^2 = 1 \]
- Expanding this gives:
\[ x^2 - 2x + 1 + y^2 - 2y + 1 = 1 \]
- Simplifying leads to:
\[ x^2 + y^2 - 2x - 2y + 1 = 0 \]

6. **Final Form**:
- Rearranging gives us the equation of the locus:
\[ x^2 + y^2 - 2x - 2y + 1 = 0 \]

### Conclusion:
The locus of the midpoint of the chord of the circle that makes an angle of \( 120^\circ \) at the center is given by the equation:
\[ x^2 + y^2 - 2x - 2y + 1 = 0 \]