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

`x^2+y^2+8x+10 y+2=0`
`x^2+y^2-5x+4y+7=0`
`2x^2+2y^2+6x+2y+1=0`
None of these

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 circles
We have two circles given by the following equations:
1. \( s_1: x^2 + y^2 + 3x + 2y + 1 = 0 \)
2. \( s_2: x^2 + y^2 + 3x + 4y + 2 = 0 \)

### Step 2: Find the equation of the common chord
The common chord of two circles can be found using the equation:
\[ s_1 - s_2 = 0 \]

Substituting the equations of the circles:
\[ (x^2 + y^2 + 3x + 2y + 1) - (x^2 + y^2 + 3x + 4y + 2) = 0 \]

### Step 3: Simplify the equation
Now, simplify the equation:
\[ x^2 + y^2 + 3x + 2y + 1 - x^2 - y^2 - 3x - 4y - 2 = 0 \]
This simplifies to:
\[ 0 + 0 + 0 - 2y - 1 = 0 \]
Thus, we have:
\[ -2y - 1 = 0 \implies -2y = 1 \implies y = -\frac{1}{2} \]

### Step 4: Determine the diameter line
The line \( y = -\frac{1}{2} \) is the line along which the diameter of the required circle lies.

### Step 5: Use the family of circles equation
The equation of the circle can be expressed as:
\[ s_1 + \lambda s_2 = 0 \]
where \( \lambda \) is a parameter.

### Step 6: Substitute the equations into the family of circles
Substituting the equations:
\[ (x^2 + y^2 + 3x + 2y + 1) + \lambda (x^2 + y^2 + 3x + 4y + 2) = 0 \]
This simplifies to:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 + (3 + 3\lambda)x + (2 + 4\lambda)y + (1 + 2\lambda) = 0 \]

### Step 7: Find the center of the circle
The center of the circle in the general form \( x^2 + y^2 + 2gx + 2fy + c = 0 \) can be identified as:
\[ g = \frac{-(3 + 3\lambda)}{2(1 + \lambda)}, \quad f = \frac{-(2 + 4\lambda)}{2(1 + \lambda)} \]

### Step 8: Set the y-coordinate of the center
Since the center lies on the line \( y = -\frac{1}{2} \):
\[ \frac{-(2 + 4\lambda)}{2(1 + \lambda)} = -\frac{1}{2} \]
Cross-multiplying gives:
\[ -(2 + 4\lambda) = -(1 + \lambda) \implies 2 + 4\lambda = 1 + \lambda \]
This simplifies to:
\[ 3\lambda = -1 \implies \lambda = -\frac{1}{3} \]

### Step 9: Substitute \(\lambda\) back into the circle equation
Now substitute \(\lambda = -\frac{1}{3}\) into the equation of the circle:
\[ (1 - \frac{1}{3})x^2 + (1 - \frac{1}{3})y^2 + (3 - 1)x + (2 - \frac{4}{3})y + (1 - \frac{2}{3}) = 0 \]
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 fractions:
\[ 2x^2 + 2y^2 + 6x + 2y + 1 = 0 \]

### Final Answer
Thus, the equation of the circle whose diameter is the common chord of the given circles is:
\[ 2x^2 + 2y^2 + 6x + 2y + 1 = 0 \]