The locus of the centre of the circle which bisects the circumferences of the circles `x^2 + y^2 = 4 & x^2 + y^2-2x + 6y + 1 =0` is :

Options

A. `2x - 6y - 15 = 0`
B. `2x +6y +15 = 0`
C. `2x - 6y +15 = 0`
D. `2x +6y - 15 = 0`

Explanation

To find the locus of the center of the circle that bisects the circumferences of the given circles, we can follow these steps:

### Step 1: Identify the equations of the circles
The given circles are:
1. \( x^2 + y^2 = 4 \)
2. \( x^2 + y^2 - 2x + 6y + 1 = 0 \)

### Step 2: Rewrite the second circle in standard form
To rewrite the second circle in standard form, we complete the square for the \(x\) and \(y\) terms.

Starting with:
\[ x^2 - 2x + y^2 + 6y + 1 = 0 \]

Completing the square for \(x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]

Completing the square for \(y\):
\[ y^2 + 6y = (y + 3)^2 - 9 \]

Substituting these into the equation:
\[ (x - 1)^2 - 1 + (y + 3)^2 - 9 + 1 = 0 \]
\[ (x - 1)^2 + (y + 3)^2 - 9 = 0 \]
\[ (x - 1)^2 + (y + 3)^2 = 9 \]

This represents a circle with center \((1, -3)\) and radius \(3\).

### Step 3: Find the common chord of the two circles
The first circle has a center at \((0, 0)\) and radius \(2\). The second circle has a center at \((1, -3)\) and radius \(3\).

The common chord of the two circles can be found by setting their equations equal to each other. The common chord can be represented as:
\[ x^2 + y^2 - 4 = 0 \quad \text{(from the first circle)} \]
\[ x^2 + y^2 - 2x + 6y + 1 = 0 \quad \text{(from the second circle)} \]

### Step 4: Set up the equation for the common chord
To find the locus of the center of the circle that bisects the circumferences of the two circles, we can express the common chord as:
\[ 2gx + 2fy + c + 4 = 0 \]
where \(c\) is a constant.

### Step 5: Substitute and simplify
From the first circle, we have:
\[ 2gx + 2fy + c + 4 = 0 \]
From the second circle, we have:
\[ 2gx + 2fy + c - 1 = 0 \]

Setting these equal gives:
\[ 2g + 1 + 2f - 3 + c - 1 = 0 \]

### Step 6: Solve for the locus
By simplifying:
\[ 2g - 6f + 15 = 0 \]

Substituting \(g = x\) and \(f = y\):
\[ 2x - 6y + 15 = 0 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ 2x - 6y = -15 \]
or
\[ x - 3y = -\frac{15}{2} \]

### Final Answer
The locus of the center of the circle which bisects the circumferences of the given circles is:
\[ x - 3y + \frac{15}{2} = 0 \]