If the equation a|z|² + ᾱz + αz̄ + d = 0 represents a circle where a, d are real constants then whi

Q: Wait, why do we reject $a=0$?

If $a=0$, the $|z|^2$ term vanishes entirely. The equation simplifies to $\bar{\alpha}z + \alpha\bar{z} + d = 0$, which is the standard form of a straight line in the complex plane, not a circle. A circle requires that quadratic term to define its curvature.

Q: Why is the radius formula $|b|^2 - c$?

This comes from completing the square in the complex plane. The expression $|z + b|^2$ expands to $(z + b)(\bar{z} + \bar{b}) = |z|^2 + \bar{b}z + b\bar{z} + |b|^2$. By comparing this to our normalized circle equation $|z|^2 + \bar{b}z + b\bar{z} + c = 0$, we see that $|z + b|^2 = |b|^2 - c$. Thus, the radius is $\sqrt{|b|^2 - c}$.

Q: What if $|\alpha|^2 - ad = 0$?

If $|\alpha|^2 - ad = 0$, the radius is zero. This represents a 'point circle' or a degenerate circle. In most JEE Advanced contexts, when we ask for the condition of a circle, we imply a non-degenerate circle with a positive radius, so we strictly require $|\alpha|^2 - ad > 0$.

Question

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

Options:

(A) $|\alpha|^2 - ad 
eq 0$

(B) $|\alpha|^2 - ad > 0$ and $a \in \mathbb{R} - \{0\}$

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

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The equation $a|z|^2 + \bar{\alpha}z + \alpha\bar{z} + d = 0$ represents a circle in the complex plane. To understand its properties, we must first examine the constraints on the coefficients.

If $a = 0$, the quadratic term $|z|^2$ vanishes, reducing the equation to a linear form. In the complex plane, a linear equation represents a straight line rather than a circle. Therefore, our first non-negotiable condition is **$a 
eq 0$**.

### The Art of Normalization

To identify the geometric parameters, we compare our equation to the standard form of a circle in the complex plane:
$$|z|^2 + \bar{b}z + b\bar{z} + c = 0$$

Since the coefficient of $|z|^2$ in the standard form is $1$, we divide our original equation by $a$:
$$|z|^2 + \frac{\bar{\alpha}}{a}z + \frac{\alpha}{a}\bar{z} + \frac{d}{a} = 0$$

By direct comparison, we identify the parameters as $b = \frac{\alpha}{a}$ and $c = \frac{d}{a}$. This mapping allows us to utilize the standard geometric properties of complex circles.

### The Radius and the Reality

The radius $r$ of a circle in the standard form $|z|^2 + \bar{b}z + b\bar{z} + c = 0$ is given by the formula:
$$r = \sqrt{|b|^2 - c}$$

Substituting our identified values for $b$ and $c$, we obtain:
$$r = \sqrt{\left|\frac{\alpha}{a}ight|^2 - \frac{d}{a}}$$

Simplifying this expression through algebraic manipulation, we find:
$$r = \sqrt{\frac{|\alpha|^2}{a^2} - \frac{d}{a}} = \sqrt{\frac{|\alpha|^2 - ad}{a^2}}$$

### Final Synthesis

For the equation to represent a real, non-degenerate circle, the radius must be a positive real number. This requires the radicand to be strictly greater than zero:
$$\frac{|\alpha|^2 - ad}{a^2} > 0$$

Since $a^2$ is always positive for any real $a 
eq 0$, the condition simplifies to the requirement that the numerator must be positive:
**$|\alpha|^2 - ad > 0$**

In conclusion, the necessary and sufficient conditions for the equation to represent a circle are **$a \in \mathbb{R} - \{0\}$** and **$|\alpha|^2 - ad > 0$**. By breaking the problem into its geometric components, we transform abstract notation into a clear, logical map.

Animated Solution for Mathematics - Complex Numbers: If the equation $a ∣ z ∣^{2} + \overset{α}{ˉ} z + α \overset{z}{ˉ} + d = 0$ represents a circle where $a, d$ are real constants then which of the following condition is correct?

Select Answer:

Visualized Solution

Analyze the Given Equation

The First Constraint: $a \neq = 0$

Normalizing the Equation

Comparing with Standard Form

Defining the Radius Formula

Substituting Values in Radius

Simplifying the Expression

Condition for a Real Circle

Final Inequality Derivation

Conclusion and Summary

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Art of Normalization

The Radius and the Reality

Final Synthesis

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