The locus of the mid-point of the chords 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)+x+y-1=0`
C. `x^(2)+y^(2)-2x-2y-1=0`
D. none of these

Answer

To find the locus of the mid-point of the chords of the circle \(x^2 + y^2 - 2x - 2y - 2 = 0\) that makes an angle of \(120^\circ\) at the center, we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 2x - 2y - 2 = 0 \]
We can rearrange this to:
\[ x^2 - 2x + y^2 - 2y = 2 \]
Now, we complete the square for both \(x\) and \(y\).

### Step 2: Complete the Square
For \(x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
For \(y\):
\[ y^2 - 2y = (y - 1)^2 - 1 \]
Substituting these back into the equation gives:
\[ (x - 1)^2 - 1 + (y - 1)^2 - 1 = 2 \]
This simplifies to:
\[ (x - 1)^2 + (y - 1)^2 = 4 \]
This represents a circle with center \(C(1, 1)\) and radius \(r = 2\).

### Step 3: Analyze the Chord and Angles
Let \(P(h, k)\) be the midpoint of the chord \(AB\) of the circle. The angle \(ACB\) is given as \(120^\circ\). The angles \(ACP\) and \(BCP\) will each be \(30^\circ\) because the total angle \(ACB\) is split into two equal angles at the midpoint \(P\).

### Step 4: Use the Sine Rule
In triangle \(ACP\), we can use the sine rule. The length \(AC\) (which is the radius) is \(2\) and the angle \(ACP\) is \(30^\circ\). We can find the length \(CP\) using:
\[ \sin(30^\circ) = \frac{CP}{AC} \Rightarrow \frac{1}{2} = \frac{CP}{2} \Rightarrow CP = 1 \]

### Step 5: Set Up the Distance Equation
Now, we can set up the distance equation from point \(C(1, 1)\) to point \(P(h, k)\):
\[ CP = \sqrt{(h - 1)^2 + (k - 1)^2} = 1 \]
Squaring both sides gives:
\[ (h - 1)^2 + (k - 1)^2 = 1 \]

### Step 6: Substitute and Rearrange
Now we can expand this equation:
\[ (h - 1)^2 + (k - 1)^2 = 1 \Rightarrow h^2 - 2h + 1 + k^2 - 2k + 1 = 1 \]
This simplifies to:
\[ h^2 + k^2 - 2h - 2k + 1 = 0 \]
Replacing \(h\) and \(k\) with \(x\) and \(y\) respectively gives:
\[ x^2 + y^2 - 2x - 2y + 1 = 0 \]

### Step 7: Final Form of the Locus
Rearranging this gives us the final equation of the locus:
\[ x^2 + y^2 - 2x - 2y + 1 = 0 \]

### Conclusion
The locus of the mid-point of the chords of the circle that makes an angle of \(120^\circ\) at the center is:
\[ x^2 + y^2 - 2x - 2y + 1 = 0 \]