The length of the common chord of the circles | vedclass

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(B) The given equations of the circles are:$C_1: x^2 + y^2 - 6x - 16 = 0$$C_2: x^2 + y^2 - 8y - 9 = 0$The equation of the common chord is given by $C_

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$5 \sqrt{3}$

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(B) The given equations of the circles are:$C_1: x^2 + y^2 - 6x - 16 = 0$$C_2: x^2 + y^2 - 8y - 9 = 0$The equation of the common chord is given by $C_1 - C_2 = 0$:$(x^2 + y^2 - 6x - 16) - (x^2 + y^2 - 8y - 9) = 0$$-6x + 8y - 7 = 0$ or $6x - 8y + 7 = 0$For circle $C_1$,the center is $(3, 0)$ and the radius $r_1 = \sqrt{3^2 + 0^2 - (-16)} = \sqrt{9 + 16} = 5$.The perpendicular distance $d$ from the center $(3, 0)$ to the common chord $6x - 8y + 7 = 0$ is:$d = \frac{|6(3) - 8(0) + 7|}{\sqrt{6^2 + (-8)^2}} = \frac{|18 + 7|}{\sqrt{36 + 64}} = \frac{25}{10} = \frac{5}{2}$The length of the common chord is $2 \sqrt{r_1^2 - d^2}$:$= 2 \sqrt{5^2 - (\frac{5}{2})^2} = 2 \sqrt{25 - \frac{25}{4}} = 2 \sqrt{\frac{100 - 25}{4}} = 2 \sqrt{\frac{75}{4}} = 2 	imes \frac{5 \sqrt{3}}{2} = 5 \sqrt{3}$

The length of the common chord of the circles $x^2 + y^2 - 6x - 16 = 0$ and $x^2 + y^2 - 8y - 9 = 0$ is:

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