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(A) The common chord of the circles $S_1: x^2+y^2+2x+3y+1=0$ and $S_2: x^2+y^2+4x+3y+2=0$ is given by $S_1 - S_2 = 0$. $(x^2+y^2+2x+3y+1) - (x^2+y^2+4

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$2x^2+2y^2+2x+6y+1=0$

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(A) The common chord of the circles $S_1: x^2+y^2+2x+3y+1=0$ and $S_2: x^2+y^2+4x+3y+2=0$ is given by $S_1 - S_2 = 0$. $(x^2+y^2+2x+3y+1) - (x^2+y^2+4x+3y+2) = 0$ $-2x - 1 = 0 \Rightarrow x = -\frac{1}{2}$. Any circle passing through the intersection of $S_1$ and $S_2$ is given by $S_1 + \lambda S_2 = 0$. $(x^2+y^2+2x+3y+1) + \lambda(x^2+y^2+4x+3y+2) = 0$ $(1+\lambda)x^2 + (1+\lambda)y^2 + (2+4\lambda)x + (3+3\lambda)y + (1+2\lambda) = 0$. Dividing by $(1+\lambda)$,the center of this circle is $(-\frac{2+4\lambda}{2(1+\lambda)}, -\frac{3+3\lambda}{2(1+\lambda)})$. Since the common chord $x = -\frac{1}{2}$ is the diameter,the center must lie on this line. $-\frac{2+4\lambda}{2(1+\lambda)} = -\frac{1}{2}$ $\Rightarrow 2+4\lambda = 1+\lambda$ $\Rightarrow 3\lambda = -1$ $\Rightarrow \lambda = -\frac{1}{3}$. Substituting $\lambda = -\frac{1}{3}$ into the equation: $(1-\frac{1}{3})x^2 + (1-\frac{1}{3})y^2 + (2-\frac{4}{3})x + (3-1)y + (1-\frac{2}{3}) = 0$ $\frac{2}{3}x^2 + \frac{2}{3}y^2 + \frac{2}{3}x + 2y + \frac{1}{3} = 0$ Multiplying by $3$,we get $2x^2+2y^2+2x+6y+1=0$.

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

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