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(B) The equation of the radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$,provided the coefficients of $x^2$ and $y^2$ a

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$7x - 3y + 18 = 0$

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(B) The equation of the radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$,provided the coefficients of $x^2$ and $y^2$ are normalized to $1$.Given circles:$S_1: x^2 + y^2 + x - y + 2 = 0$$S_2: 3x^2 + 3y^2 - 4x - 12 = 0$Divide $S_2$ by $3$ to normalize:$S_2': x^2 + y^2 - \frac{4}{3}x - 4 = 0$Now,find $S_1 - S_2' = 0$:$(x^2 + y^2 + x - y + 2) - (x^2 + y^2 - \frac{4}{3}x - 4) = 0$$x + \frac{4}{3}x - y + 2 + 4 = 0$$\frac{7}{3}x - y + 6 = 0$Multiply by $3$ to simplify:$7x - 3y + 18 = 0$

The equation of the radical axis of the circles $x^2 + y^2 + x - y + 2 = 0$ and $3x^2 + 3y^2 - 4x - 12 = 0$ is:

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