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

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(D) The given equations of the circles are:$C_1: x^2+y^2-4y=0$$C_2: x^2+y^2-8x-4y+11=0$The equation of the common chord is given by $C_1 - C_2 = 0$:$(

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

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(D) The given equations of the circles are:$C_1: x^2+y^2-4y=0$$C_2: x^2+y^2-8x-4y+11=0$The equation of the common chord is given by $C_1 - C_2 = 0$:$(x^2+y^2-4y) - (x^2+y^2-8x-4y+11) = 0$$8x - 11 = 0 \Rightarrow x = \frac{11}{8}$For circle $C_1$,the center is $O(0, 2)$ and the radius $r = \sqrt{0^2 + 2^2 - 0} = 2$.The perpendicular distance $d$ from the center $O(0, 2)$ to the chord $8x - 11 = 0$ is:$d = \frac{|8(0) - 11|}{\sqrt{8^2 + 0^2}} = \frac{11}{8}$Let $PM$ be half the length of the chord. In the right-angled triangle formed by the radius,the distance $d$,and half the chord:$PM = \sqrt{r^2 - d^2} = \sqrt{2^2 - (\frac{11}{8})^2} = \sqrt{4 - \frac{121}{64}} = \sqrt{\frac{256 - 121}{64}} = \sqrt{\frac{135}{64}} = \frac{\sqrt{135}}{8}$The length of the common chord is $2 	imes PM = 2 	imes \frac{\sqrt{135}}{8} = \frac{\sqrt{135}}{4} 	ext{ units}$.

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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