The length of the common chord of the circles | vedclass

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Question

(A) Given circles are $S_1: x^2 + y^2 + 2x + 4y - 20 = 0$ and $S_2: x^2 + y^2 + 6x - 8y + 10 = 0$.The equation of the common chord is $S_1 - S_2 = 0$.

Vedclass · Accepted Answer

$5\sqrt{\frac{3}{2}}$

Vedclass · Answer

(A) Given circles are $S_1: x^2 + y^2 + 2x + 4y - 20 = 0$ and $S_2: x^2 + y^2 + 6x - 8y + 10 = 0$.The equation of the common chord is $S_1 - S_2 = 0$.$(x^2 + y^2 + 2x + 4y - 20) - (x^2 + y^2 + 6x - 8y + 10) = 0$$-4x + 12y - 30 = 0 \Rightarrow 2x - 6y + 15 = 0$.For circle $S_1$,center $C_1 = (-1, -2)$ and radius $r_1 = \sqrt{(-1)^2 + (-2)^2 - (-20)} = \sqrt{1 + 4 + 20} = \sqrt{25} = 5$.The perpendicular distance $p$ from $C_1(-1, -2)$ to the common chord $2x - 6y + 15 = 0$ is:$p = \frac{|2(-1) - 6(-2) + 15|}{\sqrt{2^2 + (-6)^2}} = \frac{|-2 + 12 + 15|}{\sqrt{4 + 36}} = \frac{25}{\sqrt{40}} = \frac{25}{2\sqrt{10}}$.The length of the common chord is $2\sqrt{r_1^2 - p^2}$.$= 2\sqrt{25 - \frac{625}{40}} = 2\sqrt{\frac{1000 - 625}{40}} = 2\sqrt{\frac{375}{40}} = 2\sqrt{\frac{75}{8}} = 2 \cdot \frac{5\sqrt{3}}{2\sqrt{2}} = 5\sqrt{\frac{3}{2}}$.

The length of the common chord of the circles $x^2 + y^2 + 2x + 4y - 20 = 0$ and $x^2 + y^2 + 6x - 8y + 10 = 0$ is

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