The equation of the locus of the mid-points of chords of the circle `4x^(2) + 4y^(2) -12x + 4y + 1 = 0` that substend an angle `(2pi)/(3)` at its centre, is

Answer

To find the equation of the locus of the midpoints of chords of the circle given by the equation \(4x^2 + 4y^2 - 12x + 4y + 1 = 0\) that subtend an angle of \(\frac{2\pi}{3}\) at its center, we will follow these steps:

### Step 1: Convert the Circle Equation to Standard Form
First, we need to rewrite the given circle equation in the standard form.

Given:
\[ 4x^2 + 4y^2 - 12x + 4y + 1 = 0 \]
Dividing the entire equation by 4:
\[ x^2 + y^2 - 3x + y + \frac{1}{4} = 0 \]

### Step 2: Identify the Center and Radius
The general form of a circle is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From our equation, we can identify:
- \(g = -\frac{3}{2}\)
- \(f = \frac{1}{2}\)
- \(c = \frac{1}{4}\)

The center \((h, k)\) of the circle is given by:
\[ \left(-g, -f\right) = \left(\frac{3}{2}, -\frac{1}{2}\right) \]

The radius \(R\) is calculated as:
\[ R = \sqrt{g^2 + f^2 - c} = \sqrt{\left(-\frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2 - \frac{1}{4}} \]
Calculating this:
\[ R = \sqrt{\frac{9}{4} + \frac{1}{4} - \frac{1}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2} \]

### Step 3: Using the Angle Subtended by the Chord
Let \(M(h, k)\) be the midpoint of the chord \(AB\) that subtends an angle of \(\frac{2\pi}{3}\) at the center \(O\).

The angle subtended at the center is related to the distance \(OM\) (from the center to the midpoint) and the radius \(R\) by:
\[ \frac{OM}{R} = \cos\left(\frac{\theta}{2}\right) \]
Here, \(\theta = \frac{2\pi}{3}\), so \(\frac{\theta}{2} = \frac{\pi}{3}\) and \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\).

Thus,
\[ OM = R \cdot \cos\left(\frac{\pi}{3}\right) = \frac{3}{2} \cdot \frac{1}{2} = \frac{3}{4} \]

### Step 4: Expressing the Distance \(OM\)
The distance \(OM\) can be expressed using the distance formula:
\[ OM = \sqrt{\left(h - \frac{3}{2}\right)^2 + \left(k + \frac{1}{2}\right)^2} \]
Setting this equal to \(\frac{3}{4}\):
\[ \sqrt{\left(h - \frac{3}{2}\right)^2 + \left(k + \frac{1}{2}\right)^2} = \frac{3}{4} \]

### Step 5: Squaring Both Sides
Squaring both sides gives:
\[ \left(h - \frac{3}{2}\right)^2 + \left(k + \frac{1}{2}\right)^2 = \left(\frac{3}{4}\right)^2 \]
\[ \left(h - \frac{3}{2}\right)^2 + \left(k + \frac{1}{2}\right)^2 = \frac{9}{16} \]

### Step 6: Expanding the Equation
Expanding the left side:
\[ \left(h^2 - 3h + \frac{9}{4}\right) + \left(k^2 + k + \frac{1}{4}\right) = \frac{9}{16} \]
Combining terms:
\[ h^2 + k^2 - 3h + k + \frac{10}{4} = \frac{9}{16} \]
Converting \(\frac{10}{4}\) to \(\frac{40}{16}\):
\[ h^2 + k^2 - 3h + k + \frac{40}{16} - \frac{9}{16} = 0 \]
\[ h^2 + k^2 - 3h + k + \frac{31}{16} = 0 \]

### Step 7: Replacing \(h\) and \(k\) with \(x\) and \(y\)
Finally, replacing \(h\) with \(x\) and \(k\) with \(y\):
\[ x^2 + y^2 - 3x + y + \frac{31}{16} = 0 \]

### Final Answer
The equation of the locus of the midpoints of the chords is:
\[ x^2 + y^2 - 3x + y + \frac{31}{16} = 0 \]