Question

If the equation a|z|² + āz + a z̄ + d = 0 represents a circle where a, d are real constants, then which of the following condition is correct ?

• A. |a|² - a d ≥ 0 and a ∈ R
• B. |a|² - a d>0 and a ∈ R - {0}
• C. |a|² - a d ≠ 0
• D. α = 0, a, d ∈ R⁺

Hint:

The complex equation of a circle is $|z|^2 + \bar{b}z + b\bar{z} + C = 0$, where $R = \sqrt{|b|^2 - C}$.

Answer 1

The Correct Option is B

Step 1: Given $a|z|^2 + \bar{a}z + a\bar{z} + d = 0$. Since $a$ is real, $\bar{a} = a$.
Step 2: Equation becomes $a|z|^2 + az + a\bar{z} + d = 0$. Divide by $a$ (assuming $a \neq 0$): $|z|^2 + z + \bar{z} + \frac{d}{a} = 0$.
Step 3: This is of the form $|z|^2 + \bar{b}z + b\bar{z} + C = 0$ where $b=1$ and $C=d/a$.
Step 4: For this to represent a circle, the radius $R = \sqrt{|b|^2 - C}>0$.
Step 5: $|1|^2 - \frac{d}{a}>0 \implies 1>\frac{d}{a} \implies \frac{a-d}{a}>0$. Multiply by $a^2$: $a^2 - ad>0 \implies |a|^2 - ad>0$.