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10-1.Circle and System of Circles
hard
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
A
$(13, 33/4)$
B
$(33/4, -13)$
C
$(33/4, 13)$
D
None of these
Solution
(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 $\left( {\frac{{33}}{4},\,\frac{{20}}{3}} \right)$.
Standard 11
Mathematics
