A) \[|z+1+i{{|}^{2}}+|z-1-i{{|}^{2}}=8\]
B) \[|z{{|}^{2}}=|1+i{{|}^{2}}+|-1-i{{|}^{2}}\]
C) \[|z-1+i{{|}^{2}}-|z+1-i{{|}^{2}}=4\]
D) \[|z+1+i{{|}^{2}}=|z-1-i{{|}^{2}}\]
E) None of these
Correct Answer: E
Solution :
Let \[{{z}_{1}}=-1-i\] and \[{{z}_{2}}=1+i\] Then, \[{{\bar{z}}_{1}}=-1+i\] and \[{{\bar{z}}_{2}}=1-i\] Equation of circle having \[{{z}_{1}}\]and \[{{z}_{2}}\] as end points of diameter is \[(z-{{z}_{1}})\,(\bar{z}-{{\bar{z}}_{2}})+(z-{{z}_{2}})(\bar{z}-{{\bar{z}}_{1}})=0\] \[\Rightarrow \] \[z\bar{z}-z{{\bar{z}}_{2}}-{{z}_{1}}\bar{z}+{{z}_{1}}{{\bar{z}}_{2}}+z\bar{z}-z{{\bar{z}}_{1}}+{{z}_{2}}{{\bar{z}}_{1}}+{{z}_{2}}{{\bar{z}}_{1}}=0\] \[\Rightarrow \]\[2z\bar{z}-z({{\bar{z}}_{1}}+{{\bar{z}}_{2}})-\vec{z}({{z}_{1}}+{{z}_{2}})+{{z}_{1}}{{\bar{z}}_{2}}+{{z}_{2}}{{\bar{z}}_{1}}=0\] \[\Rightarrow \]\[2|z{{|}^{2}}-z(-1+i+1-i)-\bar{z}(-1-i+1+i)\] \[+{{z}_{1}}{{\bar{z}}_{2}}+{{z}_{2}}{{\bar{z}}_{1}}=0\] \[\Rightarrow \]\[2|z{{|}^{2}}-0-0+{{z}_{1}}{{\bar{z}}_{2}}+{{z}_{2}}{{\bar{z}}_{1}}=0\] \[\Rightarrow \]\[2|z{{|}^{2}}+(-1-i)(1-i)+(1+i)(-1+i)=0\] \[\Rightarrow \]\[2|z{{|}^{2}}-(-1-i)(-1-i)-(1+i)(1+i)=0\] \[\Rightarrow \] \[2|z{{|}^{2}}-|-1-i{{|}^{2}}-|1+i{{|}^{2}}=0\] \[\Rightarrow \]\[2|z{{|}^{2}}=|-1-i{{|}^{2}}+|1+i{{|}^{2}}\] Which is the required equation of circle.
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