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(B) The equation of 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) -

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

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(B) The equation of 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$$-2x-1=0 \Rightarrow 2x+1=0$.The equation of a family of circles passing through the intersection of $S_1$ and $S_2$ is $S_1 + \lambda S_2 = 0$.$(x^2+y^2+2x+3y+1) + \lambda(x^2+y^2+4x+3y+2) = 0$$(1+\lambda)x^2 + (1+\lambda)y^2 + (2+4\lambda)x + (3+3\lambda)y + (1+2\lambda) = 0$Dividing by $(1+\lambda)$,we get:$x^2+y^2 + \left(\frac{2+4\lambda}{1+\lambda}\right)x + \left(\frac{3+3\lambda}{1+\lambda}\right)y + \frac{1+2\lambda}{1+\lambda} = 0$.The center of this circle is $\left(-\frac{2+4\lambda}{2(1+\lambda)}, -\frac{3+3\lambda}{2(1+\lambda)}\right) = \left(-\frac{1+2\lambda}{1+\lambda}, -\frac{3}{2}\right)$.Since the common chord $2x+1=0$ is a diameter,the center must lie on it.$2\left(-\frac{1+2\lambda}{1+\lambda}\right) + 1 = 0$$-2(1+2\lambda) + (1+\lambda) = 0$$-2-4\lambda+1+\lambda = 0$ $\Rightarrow -3\lambda-1=0$ $\Rightarrow \lambda = -\frac{1}{3}$.Substituting $\lambda = -\frac{1}{3}$ into the family equation:$(x^2+y^2+2x+3y+1) - \frac{1}{3}(x^2+y^2+4x+3y+2) = 0$$3x^2+3y^2+6x+9y+3 - x^2-y^2-4x-3y-2 = 0$$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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