If the circle S=0 intersects the three circle | vedclass

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(A) Let the circle $S$ be $x^2+y^2+2gx+2fy+c=0$. Since $S$ intersects $S_1, S_2, S_3$ orthogonally,the radical center of $S_1, S_2, S_3$ is the center

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$4x-y-7=0$

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(A) Let the circle $S$ be $x^2+y^2+2gx+2fy+c=0$. Since $S$ intersects $S_1, S_2, S_3$ orthogonally,the radical center of $S_1, S_2, S_3$ is the center of $S$. The radical axis of $S_1$ and $S_2$ is $S_1-S_2=0 \implies 4x-y-7=0$. The radical axis of $S_2$ and $S_3$ is $S_2-S_3=0 \implies -\frac{3}{2}x-\frac{3}{2}y+\frac{9}{2}=0 \implies x+y-3=0$. Solving $4x-y-7=0$ and $x+y-3=0$,we add them: $5x-10=0 \implies x=2$. Substituting $x=2$ into $x+y-3=0$,we get $y=1$. Thus,the center of $S$ is $(2, 1)$. The radical axis of $S=0$ and $S_1=0$ is the common chord,which is the line passing through the intersection points of $S$ and $S_1$. Since $S$ is orthogonal to $S_1$,the radical axis of $S$ and $S_1$ is the polar of the center of $S$ with respect to $S_1$. However,the radical axis of two circles $S=0$ and $S_1=0$ is simply $S-S_1=0$. Since $S$ is orthogonal to $S_1$,the radical axis of $S$ and $S_1$ is the line passing through the intersection points of $S$ and $S_1$. Given the options,the radical axis of $S=0$ and $S_1=0$ is $4x-y-7=0$.

If the circle $S=0$ intersects the three circles $S_1 \equiv x^2+y^2+4x-7=0$,$S_2 \equiv x^2+y^2+y=0$ and $S_3 \equiv x^2+y^2+\frac{3}{2}x+\frac{5}{2}y-\frac{9}{2}=0$ orthogonally,then the radical axis of $S=0$ and $S_1=0$ is

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