The length of the common chord of the circles | vedclass

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(D) The equations of the circles are $S_1: x^2+y^2+3x+5y+4=0$ and $S_2: x^2+y^2+5x+3y+4=0$. The equation of the common chord is given by $S_1 - S_2 =

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$4$

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(D) The equations of the circles are $S_1: x^2+y^2+3x+5y+4=0$ and $S_2: x^2+y^2+5x+3y+4=0$. The equation of the common chord is given by $S_1 - S_2 = 0$. $(x^2+y^2+3x+5y+4) - (x^2+y^2+5x+3y+4) = 0$. $-2x + 2y = 0$,which simplifies to $y = x$. Substituting $y = x$ into $S_1$,we get $x^2 + x^2 + 3x + 5x + 4 = 0$,so $2x^2 + 8x + 4 = 0$,or $x^2 + 4x + 2 = 0$. The roots are $x = \frac{-4 \pm \sqrt{16-8}}{2} = -2 \pm \sqrt{2}$. Since $y = x$,the intersection points are $P(-2+\sqrt{2}, -2+\sqrt{2})$ and $Q(-2-\sqrt{2}, -2-\sqrt{2})$. The length of the chord $PQ$ is $\sqrt{(-2-\sqrt{2} - (-2+\sqrt{2}))^2 + (-2-\sqrt{2} - (-2+\sqrt{2}))^2} = \sqrt{(-2\sqrt{2})^2 + (-2\sqrt{2})^2} = \sqrt{8 + 8} = \sqrt{16} = 4$.

The length of the common chord of the circles $x^2+y^2+3x+5y+4=0$ and $x^2+y^2+5x+3y+4=0$ is

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