The radical centre of the circles x^2 + y^2 - | vedclass

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(D) Let the equations of the circles be:$S_1 \equiv x^2 + y^2 - 16x + 60 = 0$ $(i)$$S_2 \equiv x^2 + y^2 - 12x + 27 = 0$ $(ii)$$S_3 \equiv x^2 + y^2 -

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None of these

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(D) Let the equations of the circles be:$S_1 \equiv x^2 + y^2 - 16x + 60 = 0$ $(i)$$S_2 \equiv x^2 + y^2 - 12x + 27 = 0$ $(ii)$$S_3 \equiv x^2 + y^2 - 12y + 8 = 0$ $(iii)$The radical axis of circles $(i)$ and $(ii)$ is given by $S_1 - S_2 = 0$:$(x^2 + y^2 - 16x + 60) - (x^2 + y^2 - 12x + 27) = 0$$-4x + 33 = 0 \Rightarrow x = \frac{33}{4}$ $(iv)$The radical axis of circles $(ii)$ and $(iii)$ is given by $S_2 - S_3 = 0$:$(x^2 + y^2 - 12x + 27) - (x^2 + y^2 - 12y + 8) = 0$$-12x + 12y + 19 = 0$ $(v)$Substitute $x = \frac{33}{4}$ into equation $(v)$:$-12(\frac{33}{4}) + 12y + 19 = 0$$-99 + 12y + 19 = 0$$12y = 80 \Rightarrow y = \frac{80}{12} = \frac{20}{3}$The radical centre is $(\frac{33}{4}, \frac{20}{3})$. Since this is not among the options,the correct answer is $(d)$.

The radical centre of the circles $x^2 + y^2 - 16x + 60 = 0$,$x^2 + y^2 - 12x + 27 = 0$,and $x^2 + y^2 - 12y + 8 = 0$ is

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