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(D) Let the required circle be $S: x^2 + y^2 + 2gx + 2fy + c = 0$. If a circle $S=0$ bisects the circumference of another circle $S'=0$,then their com

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$(-2, -2)$

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(D) Let the required circle be $S: x^2 + y^2 + 2gx + 2fy + c = 0$. If a circle $S=0$ bisects the circumference of another circle $S'=0$,then their common chord passes through the centre of $S'=0$. The common chord is given by $S - S' = 0$. $1$. For $S' = x^2 + y^2 - 1 = 0$,the centre is $(0, 0)$. The common chord is $(x^2 + y^2 + 2gx + 2fy + c) - (x^2 + y^2 - 1) = 0$,which simplifies to $2gx + 2fy + c + 1 = 0$. Since it passes through $(0, 0)$,we get $c + 1 = 0$,so $c = -1$. $2$. For $S' = x^2 + y^2 + 2x - 3 = 0$,the centre is $(-1, 0)$. The common chord is $(x^2 + y^2 + 2gx + 2fy - 1) - (x^2 + y^2 + 2x - 3) = 0$,which simplifies to $2(g-1)x + 2fy + 2 = 0$. Since it passes through $(-1, 0)$,we get $2(g-1)(-1) + 2f(0) + 2 = 0$,which implies $-2g + 2 + 2 = 0$,so $2g = 4$ or $g = 2$. $3$. For $S' = x^2 + y^2 + 2y - 3 = 0$,the centre is $(0, -1)$. The common chord is $(x^2 + y^2 + 4x + 2fy - 1) - (x^2 + y^2 + 2y - 3) = 0$,which simplifies to $4x + 2(f-1)y + 2 = 0$. Since it passes through $(0, -1)$,we get $4(0) + 2(f-1)(-1) + 2 = 0$,which implies $-2f + 2 + 2 = 0$,so $2f = 4$ or $f = 2$. The centre of the required circle is $(-g, -f) = (-2, -2)$.

The coordinates of the centre of the circle which bisects the circumferences of the circles $x^2 + y^2 = 1$,$x^2 + y^2 + 2x - 3 = 0$,and $x^2 + y^2 + 2y - 3 = 0$ are

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