The equation of the circle which cuts the cir | vedclass

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(A) Let the given circles be:$S_1: x^2+y^2+4x-7=0$$S_2: x^2+y^2+\frac{3}{2}x+\frac{5}{2}y-\frac{9}{2}=0$$S_3: x^2+y^2+y=0$The centre of the circle tha

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$x^2+y^2-4x-2y-1=0$

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(A) Let the given circles be:$S_1: x^2+y^2+4x-7=0$$S_2: x^2+y^2+\frac{3}{2}x+\frac{5}{2}y-\frac{9}{2}=0$$S_3: x^2+y^2+y=0$The centre of the circle that cuts these circles orthogonally is the radical centre of $S_1, S_2,$ and $S_3$.The radical axis of $S_1$ and $S_2$ is $S_1-S_2=0$:$(4-\frac{3}{2})x - \frac{5}{2}y - 7 + \frac{9}{2} = 0$ $\Rightarrow \frac{5}{2}x - \frac{5}{2}y - \frac{5}{2} = 0$ $\Rightarrow x-y-1=0 \dots(i)$The radical axis of $S_2$ and $S_3$ is $S_2-S_3=0$:$\frac{3}{2}x + \frac{3}{2}y - \frac{9}{2} = 0 \Rightarrow x+y-3=0 \dots(ii)$Solving $(i)$ and $(ii)$,we get $2x=4 \Rightarrow x=2$ and $y=1$. The radical centre is $(2,1)$.The radius $r$ of the required circle is the length of the tangent from $(2,1)$ to $S_3$:$r^2 = 2^2+1^2+1 = 6$.The equation of the circle is $(x-2)^2+(y-1)^2 = 6$,which simplifies to $x^2+y^2-4x-2y-1=0$.

The equation of the circle which cuts the circles $x^2+y^2+4x-7=0$,$2x^2+2y^2+3x+5y-9=0$,and $x^2+y^2+y=0$ orthogonally is

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