Let C_{1} and C_{2} denote the centres of the | vedclass

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(C) The given equations of the circles are:$C_{1}: x^{2}+y^{2}=4$ (centre $C_{1} = (0,0)$,radius $r_{1} = 2$)$C_{2}: (x-2)^{2}+y^{2}=1$ (centre $C_{2}

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$7$

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(C) The given equations of the circles are:$C_{1}: x^{2}+y^{2}=4$ (centre $C_{1} = (0,0)$,radius $r_{1} = 2$)$C_{2}: (x-2)^{2}+y^{2}=1$ (centre $C_{2} = (2,0)$,radius $r_{2} = 1$)Let $N$ be the point of intersection of the common chord $PQ$ with the line joining the centres $C_{1}C_{2}$.Subtracting the two equations:$(x^{2}+y^{2}) - ((x-2)^{2}+y^{2}) = 4 - 1$$x^{2} - (x^{2}-4x+4) = 3$$4x - 4 = 3 \implies 4x = 7 \implies x = \frac{7}{4}$Since $N$ lies on the line $C_{1}C_{2}$ (the $x$-axis),the coordinate of $N$ is $(\frac{7}{4}, 0)$.The distance $C_{1}N = \frac{7}{4}$.The distance $C_{2}N = |C_{1}C_{2} - C_{1}N| = |2 - \frac{7}{4}| = \frac{1}{4}$.Both triangles $\Delta C_{1}PQ$ and $\Delta C_{2}PQ$ share the same base $PQ$.The ratio of their areas is the ratio of their heights from the base $PQ$ to the vertices $C_{1}$ and $C_{2}$ respectively,which is the ratio of the distances $C_{1}N$ and $C_{2}N$:$\frac{	ext{Area}(\Delta C_{1}PQ)}{	ext{Area}(\Delta C_{2}PQ)} = \frac{\frac{1}{2} 	imes PQ 	imes C_{1}N}{\frac{1}{2} 	imes PQ 	imes C_{2}N} = \frac{C_{1}N}{C_{2}N} = \frac{7/4}{1/4} = 7: 1$.Thus,the correct option is $C$.

Let $C_{1}$ and $C_{2}$ denote the centres of the circles $x^{2}+y^{2}=4$ and $(x-2)^{2}+y^{2}=1$ respectively and let $P$ and $Q$ be their points of intersection. Then,the areas of $\Delta C_{1} P Q$ and $\Delta C_{2} P Q$ are in the ratio (in $: 1$)

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