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(A) The equation of the family of circles passing through the intersection of $S_1: x^2+y^2+2x-2y+1=0$ and $S_2: x^2+y^2-2x+2y-2=0$ is given by $S_1 +

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$x+y=0$

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(A) The equation of the family of circles passing through the intersection of $S_1: x^2+y^2+2x-2y+1=0$ and $S_2: x^2+y^2-2x+2y-2=0$ is given by $S_1 + \lambda(S_1 - S_2) = 0$.First,find the common chord $S_1 - S_2 = 0$:$(x^2+y^2+2x-2y+1) - (x^2+y^2-2x+2y-2) = 0$$4x - 4y + 3 = 0$.Now,the family of circles is $x^2+y^2+2x-2y+1 + \lambda(4x-4y+3) = 0$,which simplifies to $x^2+y^2+(2+4\lambda)x + (-2-4\lambda)y + (1+3\lambda) = 0$.The center $(h, k)$ is given by $h = -(1+2\lambda)$ and $k = (1+2\lambda)$.Thus,$h = -k$,or $x+y=0$.The radius is $r = \sqrt{h^2+k^2-c} = \sqrt{(1+2\lambda)^2 + (1+2\lambda)^2 - (1+3\lambda)} = \sqrt{14}$.Since $h = -k$,the center always lies on the line $x+y=0$.

The centres of all circles passing through the points of intersection of the circles $x^2+y^2+2x-2y+1=0$ and $x^2+y^2-2x+2y-2=0$ and having radius $\sqrt{14}$ lie on the curve

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