The length of the common chord of the circles | vedclass

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(A) The equations of the circles are $S_1: x^2 + y^2 - 6x + 5 = 0$ and $S_2: x^2 + y^2 - 2y - 3 = 0$.The equation of the common chord is given by $S_1

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$\sqrt{6}$

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(A) The equations of the circles are $S_1: x^2 + y^2 - 6x + 5 = 0$ and $S_2: x^2 + y^2 - 2y - 3 = 0$.The equation of the common chord is given by $S_1 - S_2 = 0$.$(x^2 + y^2 - 6x + 5) - (x^2 + y^2 - 2y - 3) = 0$$-6x + 2y + 8 = 0$$3x - y - 4 = 0$ ... $(1)$For circle $S_1$,the center is $C_1(3, 0)$ and the radius $r_1 = \sqrt{3^2 + 0^2 - 5} = \sqrt{9 - 5} = 2$.The perpendicular distance $p$ from the center $(3, 0)$ to the common chord $(1)$ is:$p = \frac{|3(3) - 0 - 4|}{\sqrt{3^2 + (-1)^2}} = \frac{|9 - 4|}{\sqrt{9 + 1}} = \frac{5}{\sqrt{10}} = \frac{\sqrt{10}}{2}$.The length of the common chord is $2\sqrt{r_1^2 - p^2}$.$= 2\sqrt{2^2 - (\frac{\sqrt{10}}{2})^2} = 2\sqrt{4 - \frac{10}{4}} = 2\sqrt{\frac{16 - 10}{4}} = 2\sqrt{\frac{6}{4}} = 2 	imes \frac{\sqrt{6}}{2} = \sqrt{6}$.

The length of the common chord of the circles $x^2 + y^2 - 6x + 5 = 0$ and $x^2 + y^2 - 2y - 3 = 0$ is

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