The length of the common chord of the circles | vedclass

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(D) The equation of the common chord is given by ${S_1} - {S_2} = 0$.Subtracting the two equations: $({x^2} + {y^2} + 5x + 7y + 9) - ({x^2} + {y^2} +

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$6$

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(D) The equation of the common chord is given by ${S_1} - {S_2} = 0$.Subtracting the two equations: $({x^2} + {y^2} + 5x + 7y + 9) - ({x^2} + {y^2} + 7x + 5y + 9) = 0$.This simplifies to $-2x + 2y = 0$,or $x - y = 0$.For the circle ${x^2} + {y^2} + 5x + 7y + 9 = 0$,the center ${C_1}$ is $(-\frac{5}{2}, -\frac{7}{2})$ and the radius $r$ is $\sqrt{(-\frac{5}{2})^2 + (-\frac{7}{2})^2 - 9} = \sqrt{\frac{25}{4} + \frac{49}{4} - \frac{36}{4}} = \sqrt{\frac{38}{4}} = \sqrt{\frac{19}{2}}$.The perpendicular distance $d$ from the center ${C_1}$ to the line $x - y = 0$ is $d = \frac{|-\frac{5}{2} - (-\frac{7}{2})|}{\sqrt{1^2 + (-1)^2}} = \frac{|1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.The length of the common chord is $2\sqrt{r^2 - d^2} = 2\sqrt{\frac{19}{2} - \frac{1}{2}} = 2\sqrt{\frac{18}{2}} = 2\sqrt{9} = 2 	imes 3 = 6$.

The length of the common chord of the circles ${x^2} + {y^2} + 5x + 7y + 9 = 0$ and ${x^2} + {y^2} + 7x + 5y + 9 = 0$ is

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