The equation of the circle whose diameter is the common chord of the circles; `x^2+y^2+3x+2y+1=0` & `x^2+y^2+3x+4y+2=0` is:

Options

A. `x^2 + y^2 + 8x + 10y + 2 =0`
B. `x^2 + y^2 - 5x + 4y + 7 =0`
C. `2x^2 + 2y^2 + 6x + 2y + 1 =0`
D. none of these

Explanation

To find the equation of the circle whose diameter is the common chord of the given circles, we will follow these steps:

### Step 1: Identify the equations of the given circles
The equations of the circles are:
1. \( x^2 + y^2 + 3x + 2y + 1 = 0 \) (Circle 1)
2. \( x^2 + y^2 + 3x + 4y + 2 = 0 \) (Circle 2)

### Step 2: Find the common chord
To find the common chord of the two circles, we set the equations equal to each other:
\[ x^2 + y^2 + 3x + 2y + 1 = x^2 + y^2 + 3x + 4y + 2 \]
Subtracting the left-hand side from the right-hand side gives:
\[ 2y + 1 - 4y - 2 = 0 \]
This simplifies to:
\[ -2y - 1 = 0 \implies 2y = -1 \implies y = -\frac{1}{2} \]
Thus, the common chord is a horizontal line at \( y = -\frac{1}{2} \).

### Step 3: Use the family of circles formula
The equation of a circle can be expressed as:
\[ S + \lambda = S_1 + \lambda = 0 \]
Where \( S_1 \) is the equation of the first circle and \( \lambda \) is a parameter.

Substituting the equations of the circles:
\[ S_1: x^2 + y^2 + 3x + 2y + 1 = 0 \]
\[ S_2: x^2 + y^2 + 3x + 4y + 2 = 0 \]
We can write:
\[ S + \lambda(x^2 + y^2 + 3x + 4y + 2) = 0 \]

### Step 4: Combine the equations
Combining the two equations gives:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 + (3 + 3\lambda)x + (2 + 4\lambda)y + (1 + 2\lambda) = 0 \]

### Step 5: Find the center of the circle
The center of the circle can be found using the coefficients:
\[ \text{Center} = \left(-\frac{3 + 3\lambda}{2(1 + \lambda)}, -\frac{2 + 4\lambda}{2(1 + \lambda)}\right) \]
This center must lie on the line \( y = -\frac{1}{2} \):
\[ -\frac{2 + 4\lambda}{2(1 + \lambda)} = -\frac{1}{2} \]
Cross-multiplying gives:
\[ -2 - 4\lambda = -1 - \lambda \implies -4\lambda + \lambda = -1 + 2 \implies -3\lambda = 1 \implies \lambda = -\frac{1}{3} \]

### Step 6: Substitute \( \lambda \) back into the equation
Substituting \( \lambda = -\frac{1}{3} \) into the combined equation:
\[ (1 - \frac{1}{3})x^2 + (1 - \frac{1}{3})y^2 + (3 - 1)x + (2 - \frac{4}{3})y + (1 - \frac{2}{3}) = 0 \]
This simplifies to:
\[ \frac{2}{3}x^2 + \frac{2}{3}y^2 + 2x + \frac{2}{3}y + \frac{1}{3} = 0 \]
Multiplying through by 3 to eliminate the fractions gives:
\[ 2x^2 + 2y^2 + 6x + 2y + 1 = 0 \]

### Final Answer
Thus, the equation of the circle is:
\[ 2x^2 + 2y^2 + 6x + 2y + 1 = 0 \]