Complex Numbers And Quadratic Equations question 311
Question: The equation $ z\overline{z}+a\bar{z}+\bar{a}z+b=0,b\in R $ represents a circle if
Options:
A) $ |a{{|}^{2}}=b $
B) $ |a{{|}^{2}}>b $
C) $ |a{{|}^{2}}<b $
D) None of these
Show Answer
Answer:
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.
BETA
