The length of the common chord of the circles | vedclass

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(D) Let $S \equiv x^2+y^2+3x+5y+4=0$ and $S' \equiv x^2+y^2+5x+3y+4=0$.The equation of the common chord is $S-S'=0$.$\Rightarrow (x^2+y^2+3x+5y+4) - (

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$4$

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(D) Let $S \equiv x^2+y^2+3x+5y+4=0$ and $S' \equiv x^2+y^2+5x+3y+4=0$.The equation of the common chord is $S-S'=0$.$\Rightarrow (x^2+y^2+3x+5y+4) - (x^2+y^2+5x+3y+4) = 0$$\Rightarrow -2x+2y=0$$\Rightarrow x-y=0$.The center of $S=0$ is $C\left(-\frac{3}{2}, -\frac{5}{2}ight)$.The radius $r$ is $\sqrt{\left(\frac{3}{2}ight)^2 + \left(\frac{5}{2}ight)^2 - 4} = \sqrt{\frac{9}{4} + \frac{25}{4} - 4} = \sqrt{\frac{34}{4} - \frac{16}{4}} = \sqrt{\frac{18}{4}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}$.Let $p$ be the length of the perpendicular from $C$ to the line $x-y=0$:$p = \frac{|-\frac{3}{2} - (-\frac{5}{2})|}{\sqrt{1^2+(-1)^2}} = \frac{|1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.The length of the common chord $AB = 2\sqrt{r^2-p^2}$.$AB = 2\sqrt{\frac{9}{2} - \frac{1}{2}} = 2\sqrt{\frac{8}{2}} = 2\sqrt{4} = 2 	imes 2 = 4$.

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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