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

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Question

(d) ${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

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(d) ${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 circle $(i)$ and circle $(ii)$ is

${S_1} - {S_2} = 0\, \Rightarrow \, - 4x + 33 = 0$ ....$(iv)$

the radical axis of circle $(ii)$ and circle $(iii)$ is ${S_2} - {S_3} = 0$

$ \Rightarrow \,\, - 12 + 12y + 19 = 0$ .....$(v)$

Solving $(iv)$ and $(v),$ we get the radical centre $\left( {\frac{{33}}{4},\,\frac{{20}}{3}} \right)$.

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

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