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

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(A) Given equations of circles are $S_1: x^2+y^2-4x-8y+4=0$ and $S_2: x^2+y^2-8x-12y+16=0$.Equation of the common chord is $S_1-S_2=0$.$(x^2+y^2-4x-8y

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$\sqrt{46}$

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(A) Given equations of circles are $S_1: x^2+y^2-4x-8y+4=0$ and $S_2: x^2+y^2-8x-12y+16=0$.Equation of the common chord is $S_1-S_2=0$.$(x^2+y^2-4x-8y+4) - (x^2+y^2-8x-12y+16) = 0$$4x+4y-12=0 \implies x+y-3=0$.For $S_1: x^2-4x+y^2-8y+4=0 \implies (x-2)^2+(y-4)^2 = 4+16-4 = 16$.Centre $O(2, 4)$,radius $r_1 = 4$.The perpendicular distance $d$ from $O(2, 4)$ to the line $x+y-3=0$ is:$d = \frac{|2+4-3|}{\sqrt{1^2+1^2}} = \frac{3}{\sqrt{2}}$.Let $L$ be the length of the common chord. Then $L = 2\sqrt{r_1^2-d^2}$.$L = 2\sqrt{4^2 - (\frac{3}{\sqrt{2}})^2} = 2\sqrt{16 - \frac{9}{2}} = 2\sqrt{\frac{32-9}{2}} = 2\sqrt{\frac{23}{2}} = \sqrt{4 	imes \frac{23}{2}} = \sqrt{46}$.Thus,the correct option is $A$.

The length of the common chord of the two circles $x^2+y^2-4x-8y+4=0$ and $x^2+y^2-8x-12y+16=0$ is

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