A) \[\left( \frac{1}{7},\frac{9}{7} \right)\]
B) \[\left( -\frac{1}{7},-\frac{9}{7} \right)\]
C) \[\left( \frac{1}{7},-\frac{9}{7} \right)\]
D) \[\left( -\frac{1}{7},\frac{9}{7} \right)\]
Correct Answer: D
Solution :
The given circles Let \[{{S}_{1}}\equiv {{x}^{2}}+{{y}^{2}}+6x-1=0,\] Centre \[{{C}_{1}}=(-3,0)\] Radius \[{{R}_{1}}=\sqrt{10}\] \[{{S}_{2}}\equiv {{x}^{2}}+{{y}^{2}}-3y+2=0,\] \[{{C}_{2}}=(0,3/2),\,\,\,{{R}_{2}}=1/2\] \[{{S}_{3}}\equiv {{x}^{2}}+{{y}^{2}}+x+y-3=0,\] \[{{C}_{3}}=(-1/2,\,\,-1/2),\,\,{{R}_{3}}=\sqrt{7/2}\] Let \[S\equiv {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]be equation of circle which cut orthogonal all three circles. Then, by condition of orthogonality \[-2g\,(-3)+2(-f)\,(0)\,c-1\] \[\Rightarrow \] \[6g=c-1\] ...(i) \[-2g(0)+2f(-3/2)=c+2\] \[\Rightarrow \] \[-3f=c+2\] ...(ii) \[-2g(-1/2)+2f(1/2)=c-3\] \[\Rightarrow \] \[g+f=c-3\] ?..(iii) On solving Eqs. (i), (ii) and (iii), we get \[g=1/7,\,\,f=-9/7\] So, centre is \[(-g,-f)=(-1/7,9/7)\]
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