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

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(D) The given equations of the circles are: $x^2+y^2-4y=0$  $(1)$ $x^2+y^2-8x-4y+11=0$  $(2)$ Subtracting equation $(1)$ from $(2)$,we get the equatio

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$\frac{\sqrt{135}}{4} 	ext{ units}$

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(D) The given equations of the circles are: $x^2+y^2-4y=0$  $(1)$ $x^2+y^2-8x-4y+11=0$  $(2)$ Subtracting equation $(1)$ from $(2)$,we get the equation of the common chord: $(x^2+y^2-8x-4y+11) - (x^2+y^2-4y) = 0$ $-8x+11=0$ $\Rightarrow 8x=11$ $\Rightarrow x=\frac{11}{8}$ For the first circle $x^2+y^2-4y=0$,the center is $C(0, 2)$ and the radius $r = \sqrt{0^2 + 2^2 - 0} = 2$. The perpendicular distance $d$ from the center $(0, 2)$ to the line $x = \frac{11}{8}$ is: $d = |0 - \frac{11}{8}| = \frac{11}{8}$ Let $L$ be the length of the common chord. The formula for the length of the chord is $L = 2\sqrt{r^2 - d^2}$. $L = 2\sqrt{2^2 - (\frac{11}{8})^2} = 2\sqrt{4 - \frac{121}{64}}$ $L = 2\sqrt{\frac{256 - 121}{64}} = 2\sqrt{\frac{135}{64}} = 2 	imes \frac{\sqrt{135}}{8} = \frac{\sqrt{135}}{4} 	ext{ units}$.

The length of the common chord of the two circles $x^2+y^2-4y=0$ and $x^2+y^2-8x-4y+11=0$ is

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