The coordinates of the radical centre of the | vedclass

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(D) Let the equations of the circles be:$S_1: x^2 + y^2 - 4x - 2y + 6 = 0$$S_2: x^2 + y^2 - 2x - 4y - 1 = 0$$S_3: x^2 + y^2 - 12x + 2y + 30 = 0$The ra

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None of these

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(D) Let the equations of the circles be:$S_1: x^2 + y^2 - 4x - 2y + 6 = 0$$S_2: x^2 + y^2 - 2x - 4y - 1 = 0$$S_3: x^2 + y^2 - 12x + 2y + 30 = 0$The radical axis of $S_1$ and $S_2$ is given by $S_1 - S_2 = 0$:$(x^2 + y^2 - 4x - 2y + 6) - (x^2 + y^2 - 2x - 4y - 1) = 0$$-2x + 2y + 7 = 0$ or $2x - 2y = 7$ $(i)$The radical axis of $S_2$ and $S_3$ is given by $S_2 - S_3 = 0$:$(x^2 + y^2 - 2x - 4y - 1) - (x^2 + y^2 - 12x + 2y + 30) = 0$$10x - 6y - 31 = 0$ $(ii)$Solving equations $(i)$ and $(ii)$:From $(i)$,$2x = 2y + 7 \implies x = y + 3.5$Substitute into $(ii)$:$10(y + 3.5) - 6y - 31 = 0$$10y + 35 - 6y - 31 = 0$$4y + 4 = 0 \implies y = -1$$x = -1 + 3.5 = 2.5$The radical centre is $(2.5, -1)$,which is not among the given options.

The coordinates of the radical centre of the three circles $x^2 + y^2 - 4x - 2y + 6 = 0$,$x^2 + y^2 - 2x - 4y - 1 = 0$,and $x^2 + y^2 - 12x + 2y + 30 = 0$ are

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