(C) The radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$. For the first two circles: $(x^2 + y^2 + 2gx - 4y + 4) - (x^2

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(C) The radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$. For the first two circles: $(x^2 + y^2 + 2gx - 4y + 4) - (x^2 + y^2 + 6x + 2fy + 12) = 0$ $(2g - 6)x - (4 + 2f)y - 8 = 0$. Since $(-1, -1)$ is the radical centre,it must satisfy the equation of the radical axis: $(2g - 6)(-1) - (4 + 2f)(-1) - 8 = 0$ $-2g + 6 + 4 + 2f - 8 = 0$ $-2g + 2f + 2 = 0$ $2f - 2g = -2$ $g - f = 1$.

Class 11 Mathematics 10-1.Circle and System of Circles System of circles

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If $(-1, -1)$ is the radical centre of the circles $x^2 + y^2 + 2gx - 4y + 4 = 0$,$x^2 + y^2 + 6x + 2fy + 12 = 0$,and $x^2 + y^2 + 10y + 20 = 0$,then $g - f = $

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If $(-1, -1)$ is the radical centre of the circles $x^2 + y^2 + 2gx - 4y + 4 = 0$,$x^2 + y^2 + 6x + 2fy + 12 = 0$,and $x^2 + y^2 + 10y + 20 = 0$,then $g - f = $

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