A) \[|a{{|}^{2}}=b\]
B) \[|a{{|}^{2}}>b\]
C) \[|a{{|}^{2}}<b\]
D) None of these
Correct Answer: B
Solution :
By adding \[a\overline{a}\] on both the sides of \[z\overline{z}+a\overline{z}+\overline{a}z=-b\] we get, \[(z+a)(\overline{z}+\overline{a})=a\overline{a}-b\] \[|z+a{{|}^{2}}=\,|a{{|}^{2}}-b,\{\because z\overline{z}=|z{{|}^{2}}\}\] This equation will represent a circle with centre \[z=-a,\]if \[|a{{|}^{2}}-b>0,i.e.|a{{|}^{2}}>b\] since \[|a{{|}^{2}}=b\] represents point circle only.
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