The length of the common chord of the circles | vedclass

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(A) The equations of the circles are $S_1: x^2 + y^2 - 12 = 0$ and $S_2: x^2 + y^2 - 4x + 3y - 2 = 0$.The equation of the common chord is given by $S_

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$4$

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(A) The equations of the circles are $S_1: x^2 + y^2 - 12 = 0$ and $S_2: x^2 + y^2 - 4x + 3y - 2 = 0$.The equation of the common chord is given by $S_1 - S_2 = 0$.$(x^2 + y^2 - 12) - (x^2 + y^2 - 4x + 3y - 2) = 0$$4x - 3y - 10 = 0$.The center of the first circle $S_1$ is $C_1(0, 0)$ and its radius $R_1 = \sqrt{12} = 2\sqrt{3}$.The perpendicular distance $p_1$ from the center $(0, 0)$ to the line $4x - 3y - 10 = 0$ is:$p_1 = \frac{|4(0) - 3(0) - 10|}{\sqrt{4^2 + (-3)^2}} = \frac{|-10|}{\sqrt{16 + 9}} = \frac{10}{5} = 2$.The length of the common chord is $2\sqrt{R_1^2 - p_1^2}$.Length $= 2\sqrt{12 - 2^2} = 2\sqrt{12 - 4} = 2\sqrt{8} = 2(2\sqrt{2}) = 4\sqrt{2}$.

The length of the common chord of the circles $x^2 + y^2 = 12$ and $x^2 + y^2 - 4x + 3y - 2 = 0$ is (in $\sqrt{2}$)

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