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(A) The equation of the common chord of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.First,write the equations in the form $x^2 + y

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$10x - 3y - 18 = 0$

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(A) The equation of the common chord of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.First,write the equations in the form $x^2 + y^2 + 2gx + 2fy + c = 0$.The first circle is $3x^2 + 3y^2 - 2x + 12y - 9 = 0$. Dividing by $3$,we get $x^2 + y^2 - \frac{2}{3}x + 4y - 3 = 0$ $(S_1)$.The second circle is $x^2 + y^2 + 6x + 2y - 15 = 0$ $(S_2)$.The common chord is $S_1 - S_2 = 0$:$(x^2 + y^2 - \frac{2}{3}x + 4y - 3) - (x^2 + y^2 + 6x + 2y - 15) = 0$$(-\frac{2}{3} - 6)x + (4 - 2)y + (-3 + 15) = 0$$-\frac{20}{3}x + 2y + 12 = 0$Multiplying by $-3$,we get $20x - 6y - 36 = 0$,which simplifies to $10x - 3y - 18 = 0$.

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

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