A) \[~5{{x}^{2}}+\text{ }5{{y}^{2}}-\text{ }60x\text{ }+\text{ }7\text{ }=\text{ }0\]
B) \[~5{{x}^{2}}+\text{ }5{{y}^{2}}\text{+ }60x\text{ }+\text{ }7\text{ }=\text{ }0\]
C) \[~5{{x}^{2}}+\text{ }5{{y}^{2}}\text{+ }60x\text{ }+\text{ 12 }=\text{ }0\]
D) None of the above
Correct Answer: B
Solution :
Let \[{{S}_{1}}={{x}^{2}}+{{y}^{2}}+4x+3=0\] and \[{{S}_{2}}={{x}^{2}}+{{y}^{2}}-6x+5=0\] Given , \[\sqrt{\frac{{{S}_{1}}}{{{S}_{2}}}}=\frac{2}{3}\] \[\Rightarrow \] \[\frac{\sqrt{x_{1}^{2}+y_{1}^{2}+4{{x}_{1}}+3}}{\sqrt{x_{1}^{2}+y_{1}^{2}-6{{x}_{1}}+5}}=\frac{2}{3}\] \[\Rightarrow \] \[9x_{1}^{2}+9y_{1}^{2}+36{{x}_{1}}+27\] \[-4x_{1}^{2}-4y_{1}^{2}+24{{x}_{1}}-20=0\] \[\Rightarrow \] \[5x_{1}^{2}+5y_{1}^{2}+60{{x}_{1}}+7=0\] Hence, locus of a point is \[5{{x}^{2}}+5{{y}^{2}}+60x+7=0\]
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