A) \[2\]
B) \[1\]
C) \[5\]
D) \[4\]
Correct Answer: C
Solution :
Given, equation of circles are \[{{S}_{1}}\equiv {{x}^{2}}+{{y}^{2}}-6x-6y+4=0\] \[{{S}_{2}}\equiv {{x}^{2}}+{{y}^{2}}-2x-4y+3=0\] and \[{{S}_{3}}\equiv {{x}^{2}}+{{y}^{2}}+2kx+2y+1=0\] Now, radical axis of circle \[{{S}_{1}}\] and \[{{S}_{2}}\] is \[{{S}_{1}}-{{S}_{2}}=0\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-6x-6y+4-{{x}^{2}}-{{y}^{2}}+2x\] \[+4y-3=0\] \[\Rightarrow \] \[-4x-2y+1=0\] \[\Rightarrow \] \[4x+2y+1=0\] ??(i) Radical axis of circle \[{{S}_{2}}\] and \[{{S}_{3}}\]is \[{{S}_{2}}-{{S}_{3}}=0\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}-2x-4y+3-{{x}^{2}}-{{y}^{2}}-2kx\] \[-2y-1=0\] \[\Rightarrow \] \[-(2+2k)x-6y+2=0\] \[\Rightarrow \] \[(2+2k)x+6y-2=0\] ?..(ii) For existence of radical centre \[\left| \begin{matrix} 2 & 2 \\ 2+2k & 6 \\ \end{matrix} \right|\ne 0\] \[\Rightarrow \] \[24-2(2+2k)\ne 0\] \[\Rightarrow \] \[24-4-4\,\,k\ne 0\] \[\Rightarrow \] \[20-4\,\,k\ne 0\] \[\Rightarrow \] \[k\ne 5\]
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