The equation of the line passing through the intersection of the circles `3x^(2) +3y^(2) -2x +12y -9=0` and `x^(2)+y^(2)+6x+2y-15=0` is

Options

A. `8x-10y-7=0`
B. `10x-3y-27=0`
C. `4x+10y-5=0`
D. none

Answer

To find the equation of the line passing through the intersection of the given circles, we will follow these steps:

### Step 1: Write down the equations of the circles
The equations of the circles are given as:
1. \( S_1: 3x^2 + 3y^2 - 2x + 12y - 9 = 0 \)
2. \( S_2: x^2 + y^2 + 6x + 2y - 15 = 0 \)

### Step 2: Simplify the first circle's equation
We can simplify the first circle's equation by dividing all terms by 3:
\[ S_1: x^2 + y^2 - \frac{2}{3}x + 4y - 3 = 0 \]

### Step 3: Write the equations in standard form
Now we have:
1. \( S_1: x^2 + y^2 - \frac{2}{3}x + 4y - 3 = 0 \)
2. \( S_2: x^2 + y^2 + 6x + 2y - 15 = 0 \)

### Step 4: Find the equation of the common chord
To find the equation of the line passing through the intersection of the two circles, we can use the formula \( S_1 - S_2 = 0 \).

Subtract \( S_2 \) from \( S_1 \):
\[ S_1 - S_2 = \left( x^2 + y^2 - \frac{2}{3}x + 4y - 3 \right) - \left( x^2 + y^2 + 6x + 2y - 15 \right) = 0 \]

### Step 5: Simplify the equation
The \( x^2 \) and \( y^2 \) terms cancel out:
\[ -\frac{2}{3}x + 4y - 3 - 6x - 2y + 15 = 0 \]
Combine like terms:
\[ -\frac{2}{3}x - 6x + 4y - 2y + 12 = 0 \]
This simplifies to:
\[ -\frac{20}{3}x + 2y + 12 = 0 \]

### Step 6: Clear the fractions
To eliminate the fraction, multiply the entire equation by 3:
\[ -20x + 6y + 36 = 0 \]
Rearranging gives:
\[ 20x - 6y - 36 = 0 \]

### Step 7: Final equation
We can simplify this further by dividing the entire equation by 2:
\[ 10x - 3y - 18 = 0 \]

Thus, the equation of the line passing through the intersection of the circles is:
\[ 10x - 3y - 18 = 0 \]