The length (in units) of the common chord of | vedclass

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(B) The equation of the common chord of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.Given circles are $S_1: x^2+y^2+2x+3y+1=0$ and

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$2\sqrt{2}$

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(B) The equation of the common chord of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.Given circles are $S_1: x^2+y^2+2x+3y+1=0$ and $S_2: x^2+y^2+4x+3y+2=0$.Subtracting $S_2$ from $S_1$,we get $(x^2+y^2+2x+3y+1) - (x^2+y^2+4x+3y+2) = 0$,which simplifies to $-2x - 1 = 0$,or $x = -\frac{1}{2}$.Substitute $x = -\frac{1}{2}$ into the first circle equation: $(-\frac{1}{2})^2 + y^2 + 2(-\frac{1}{2}) + 3y + 1 = 0$.$\frac{1}{4} + y^2 - 1 + 3y + 1 = 0 \Rightarrow y^2 + 3y + \frac{1}{4} = 0$.Multiplying by $4$,we get $4y^2 + 12y + 1 = 0$.Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $y = \frac{-12 \pm \sqrt{144 - 16}}{8} = \frac{-12 \pm \sqrt{128}}{8} = \frac{-12 \pm 8\sqrt{2}}{8} = -\frac{3}{2} \pm \sqrt{2}$.The endpoints of the common chord are $(-\frac{1}{2}, -\frac{3}{2} + \sqrt{2})$ and $(-\frac{1}{2}, -\frac{3}{2} - \sqrt{2})$.The length of the common chord is the distance between these points: $\sqrt{(-\frac{1}{2} - (-\frac{1}{2}))^2 + ((-\frac{3}{2} + \sqrt{2}) - (-\frac{3}{2} - \sqrt{2}))^2} = \sqrt{0^2 + (2\sqrt{2})^2} = 2\sqrt{2}$ units.

The length (in units) of 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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