The length of the common chord of the circles | vedclass

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(C) The equation of the common chord is obtained by subtracting the two circle equations: $(x^2+y^2-6x-4y+9) - (x^2+y^2-8x-6y+23) = 0$. This simplifie

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$2\sqrt{2}$

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(C) The equation of the common chord is obtained by subtracting the two circle equations: $(x^2+y^2-6x-4y+9) - (x^2+y^2-8x-6y+23) = 0$. This simplifies to $2x + 2y - 14 = 0$,or $x + y - 7 = 0$. For the circle $x^2+y^2-6x-4y+9=0$,the center is $(3, 2)$ and the radius $r = \sqrt{3^2 + 2^2 - 9} = \sqrt{9+4-9} = 2$. The perpendicular distance $d$ from the center $(3, 2)$ to the line $x + y - 7 = 0$ is $d = \frac{|3 + 2 - 7|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \sqrt{2}$. The length of the common chord is $2\sqrt{r^2 - d^2} = 2\sqrt{2^2 - (\sqrt{2})^2} = 2\sqrt{4 - 2} = 2\sqrt{2}$.

The length of the common chord of the circles $x^2+y^2-6x-4y+9=0$ and $x^2+y^2-8x-6y+23=0$ is

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