The length of the common chord of the two cir | vedclass

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(A) The equations of the circles are:$S_1: x^2 + y^2 + \frac{7}{2}x - \frac{5}{2}y + 1 = 0$ (Dividing by $2$)$S_2: x^2 + y^2 - 4x + 8y - 18 = 0$The eq

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$2 \sqrt{\frac{1102}{333}}$

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(A) The equations of the circles are:$S_1: x^2 + y^2 + \frac{7}{2}x - \frac{5}{2}y + 1 = 0$ (Dividing by $2$)$S_2: x^2 + y^2 - 4x + 8y - 18 = 0$The equation of the common chord is $S_1 - S_2 = 0$:$(x^2 + y^2 + \frac{7}{2}x - \frac{5}{2}y + 1) - (x^2 + y^2 - 4x + 8y - 18) = 0$$\frac{15}{2}x - \frac{21}{2}y + 19 = 0 \Rightarrow 15x - 21y + 38 = 0$The center of circle $S_2$ is $C_2(2, -4)$ and its radius $r = \sqrt{2^2 + (-4)^2 - (-18)} = \sqrt{4 + 16 + 18} = \sqrt{38}$.The perpendicular distance $d$ from $C_2(2, -4)$ to the line $15x - 21y + 38 = 0$ is:$d = \frac{|15(2) - 21(-4) + 38|}{\sqrt{15^2 + (-21)^2}} = \frac{|30 + 84 + 38|}{\sqrt{225 + 441}} = \frac{152}{\sqrt{666}}$.The length of the common chord is $2\sqrt{r^2 - d^2}$:$2\sqrt{38 - \frac{152^2}{666}} = 2\sqrt{38 - \frac{23104}{666}} = 2\sqrt{\frac{25308 - 23104}{666}} = 2\sqrt{\frac{2204}{666}} = 2\sqrt{\frac{1102}{333}}$.

The length of the common chord of the two circles $2x^{2} + 2y^{2} + 7x - 5y + 2 = 0$ and $x^{2} + y^{2} - 4x + 8y - 18 = 0$ is:

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