The locus of the mid-points of the chords of the circle `x^2+ y^2-2x-4y - 11=0` which subtends an angle of `60^@` at center is

Options

A. `( x-1)^(2) +( y-2)^(2) =12`
B. `( x+1)^(2) +( y+2)^(2) =12`
C. `( x+2)^(2)+ ( y-1)^(2) =12 `
D. `(x-1)^(2) +( y-2)^(2) =18 `

Answer

To find the locus of the midpoints of the chords of the circle \( x^2 + y^2 - 2x - 4y - 11 = 0 \) that subtend an angle of \( 60^\circ \) at the center, we can follow these steps:

### Step 1: Rewrite the equation of the circle
We start with the given equation of the circle:
\[ x^2 + y^2 - 2x - 4y - 11 = 0 \]
We can rearrange this equation by completing the square.

### Step 2: Complete the square
For \( x \):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
For \( y \):
\[ y^2 - 4y = (y - 2)^2 - 4 \]
Now substituting back into the equation:
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 - 11 = 0 \]
This simplifies to:
\[ (x - 1)^2 + (y - 2)^2 - 16 = 0 \]
Thus, we have:
\[ (x - 1)^2 + (y - 2)^2 = 16 \]
This indicates that the center of the circle is \( (1, 2) \) and the radius \( r = 4 \).

### Step 3: Determine the relationship involving the angle subtended
Let \( O \) be the center of the circle, and let \( M \) be the midpoint of a chord \( AB \) that subtends an angle of \( 60^\circ \) at \( O \). The angle subtended at the center by the chord can be related to the distance from the center \( O \) to the midpoint \( M \).

### Step 4: Use the cosine rule
In triangle \( OMB \), we can use the cosine of the angle:
\[ \cos(30^\circ) = \frac{MO}{OB} \]
Here, \( OB \) is the radius \( r = 4 \), and \( MO \) is the distance from the center to the midpoint \( M \).

### Step 5: Substitute values
We know that \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ \frac{\sqrt{3}}{2} = \frac{MO}{4} \]
Thus, we can solve for \( MO \):
\[ MO = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} \]

### Step 6: Use the distance formula
The distance \( MO \) can also be expressed using the coordinates of \( M(h, k) \) and \( O(1, 2) \):
\[ MO = \sqrt{(h - 1)^2 + (k - 2)^2} \]
Setting this equal to \( 2\sqrt{3} \):
\[ \sqrt{(h - 1)^2 + (k - 2)^2} = 2\sqrt{3} \]

### Step 7: Square both sides
Squaring both sides gives:
\[ (h - 1)^2 + (k - 2)^2 = (2\sqrt{3})^2 = 12 \]

### Step 8: Substitute \( h \) and \( k \)
Let \( h = x \) and \( k = y \):
\[ (x - 1)^2 + (y - 2)^2 = 12 \]

### Final Result
The equation of the locus of the midpoints of the chords is:
\[ (x - 1)^2 + (y - 2)^2 = 12 \]