The correct option is B |a|2>b
By adding a¯a on both the sides of z¯z+a¯z+¯az=−b
we get, (z+a)(¯z+¯a)=a¯a−b
⇒ |z+a|2=|a|2−b, ∵ z¯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.