Find the equation of the radical axis of circles `x^2+y^2+x-y+2=0` and `3x^2+3y^2-4x-12=0`

Options

A. `2x^(2)+2y^(2)-5x+y-14=0`
B. `7x-3y+18=0`
C. `5x-y+14=0`
D. None of these

Answer

To find the equation of the radical axis of the circles given by the equations \( x^2 + y^2 + x - y + 2 = 0 \) and \( 3x^2 + 3y^2 - 4x - 12 = 0 \), we can follow these steps:

### Step 1: Write the equations of the circles in standard form
The first circle is already in the standard form:
\[ s_1: x^2 + y^2 + x - y + 2 = 0 \]

For the second circle, we need to divide the entire equation by 3 to simplify it:
\[ s_2: 3x^2 + 3y^2 - 4x - 12 = 0 \implies x^2 + y^2 - \frac{4}{3}x - 4 = 0 \]

### Step 2: Set up the equations for the radical axis
The radical axis can be found by equating \( s_1 \) and \( s_2 \):
\[ s_1 = s_2 \]
Substituting the equations:
\[ x^2 + y^2 + x - y + 2 = x^2 + y^2 - \frac{4}{3}x - 4 \]

### Step 3: Simplify the equation
Cancel out \( x^2 + y^2 \) from both sides:
\[ x - y + 2 = -\frac{4}{3}x - 4 \]
Now, rearranging the terms gives:
\[ x + \frac{4}{3}x - y + 2 + 4 = 0 \]
Combine like terms:
\[ \left(1 + \frac{4}{3}\right)x - y + 6 = 0 \]
This simplifies to:
\[ \frac{7}{3}x - y + 6 = 0 \]

### Step 4: Rearranging to standard linear form
To express this in a more standard linear form:
\[ y = \frac{7}{3}x + 6 \]

### Final Equation of the Radical Axis
Thus, the equation of the radical axis of the two circles is:
\[ \frac{7}{3}x - y + 6 = 0 \]