If the equation $\mathrm{a}|\mathrm{z}|^{2}+\overline{\bar{\alpha} \mathrm{z}+\alpha \overline{\mathrm{z}}}+\mathrm{d}=0$ represents a circle where $\mathrm{a}, \mathrm{d}$ are real constants, then which of the following condition is correct?

(D) $|\alpha|^{2}-a d>0$ and $\mathrm{a} \in R-\{0\}$

(A) $|\alpha|^{2}-\mathrm{ad} \neq 0$

(B) $\alpha=0, a, d \in R^{+}$

(C) $|\alpha|^{2}-\mathrm{ad} \geq 0$ and $\mathrm{a} \in \mathrm{R}$

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Complex Numbers

If the equation

a ∣ z ∣^{2} + \overline{\overset{α}{ˉ} z + α \overline{z}} + d = 0

represents a circle where

a, d

are real constants, then which of the following condition is correct?

Answer: (D)

Write

α = p + i q

and

z = x + i y

. Then

\overset{α}{ˉ} z + α \overset{z}{ˉ} = 2 (p x + q y)

is real, so the outer conjugation leaves it unchanged and the equation reads

a (x^{2} + y^{2}) + 2 p x + 2 q y + d = 0

A circle needs the squared terms to survive, so

a \neq = 0

, and dividing by

a

puts the equation in the form

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

with

g = \frac{p}{a}

f = \frac{q}{a}

and

c = \frac{d}{a}

. Its radius satisfies

r^{2} = g^{2} + f^{2} - c = \frac{p ^{2} + q ^{2}}{a ^{2}} - \frac{d}{a} = \frac{∣ α ∣ ^{2} - a d}{a ^{2}}

A circle of positive radius therefore exists exactly when

∣ α ∣^{2} - a d > 0

and

a \in R - {0}

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Single correct option