A) \[3{{x}^{2}}+\text{ }3{{y}^{2}}+10y\text{ }-3=0\]
B) \[~3{{x}^{2}}+3{{y}^{2}}+10y+3=0\]
C) \[3{{x}^{2}}-3{{y}^{2}}-10y-3=0\]
D) \[{{x}^{2}}+{{y}^{2}}-5y+3=0\]
Correct Answer: B
Solution :
Since, \[\left| \frac{z-i}{z+i} \right|=2\] and \[z=x+iy\] \[\therefore \] \[\left| \frac{x+iy-1}{x+iy+i} \right|=2\] \[\Rightarrow \] \[\left| \frac{x+(y-1)i}{x+(y+1)i} \right|=2\] \[\Rightarrow \] \[|x+i(y-1)|=2|x+(y+1)i|\] \[\Rightarrow \]\[{{x}^{2}}+{{(y-1)}^{2}}=4({{x}^{2}}+{{(y+1)}^{2}})\] \[\Rightarrow \,\,{{x}^{2}}+{{y}^{2}}-2y+1=4{{x}^{2}}+4{{y}^{2}}+8y+4\] \[\Rightarrow \]\[3{{x}^{2}}+3{{y}^{2}}+10y+3=0\]
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