The length of the common chord of the circles | vedclass

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(B) The equations of the circles are $C_1: x^2+y^2-6x+5=0$ and $C_2: x^2+y^2+4y-5=0$. Subtracting the two equations gives the equation of the common c

Vedclass · Accepted Answer

$\frac{12}{\sqrt{13}}$

Vedclass · Answer

(B) The equations of the circles are $C_1: x^2+y^2-6x+5=0$ and $C_2: x^2+y^2+4y-5=0$. Subtracting the two equations gives the equation of the common chord: $(x^2+y^2-6x+5) - (x^2+y^2+4y-5) = 0$,which simplifies to $-6x-4y+10=0$,or $3x+2y-5=0$. The center of $C_1$ is $(3, 0)$ and its radius $r_1 = \sqrt{3^2+0^2-5} = \sqrt{4} = 2$. The distance $d$ from the center $(3, 0)$ to the line $3x+2y-5=0$ is $d = \frac{|3(3)+2(0)-5|}{\sqrt{3^2+2^2}} = \frac{|9-5|}{\sqrt{13}} = \frac{4}{\sqrt{13}}$. The length of the common chord is $2\sqrt{r_1^2-d^2} = 2\sqrt{2^2 - (\frac{4}{\sqrt{13}})^2} = 2\sqrt{4 - \frac{16}{13}} = 2\sqrt{\frac{52-16}{13}} = 2\sqrt{\frac{36}{13}} = \frac{12}{\sqrt{13}}$.

The length of the common chord of the circles $x^2+y^2-6x+5=0$ and $x^2+y^2+4y-5=0$ is:

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