For α, β, z ∈ C and λ 1, if √(λ - 1) is the radius of the circle |z - α|² + |z - β|² = 2λ, then |α

Q: Why can't we just use the distance formula in Cartesian coordinates?

You absolutely could, but it would be a nightmare of algebra! Using complex numbers and the property $|w|^2 = w\bar{w}$ keeps the geometry elegant and avoids the messy expansion of $(x-a_1)^2 + (y-a_2)^2$. Trust the complex algebra; it is designed to simplify these exact types of problems.

Q: Wait, why do we reject the negative root for the distance?

Great catch! In the final step, we found $|\alpha - \beta|^2 = 4$. Mathematically, this implies $|\alpha - \beta| = \pm 2$. However, $|\alpha - \beta|$ represents the distance between two points in the complex plane, which is by definition a non-negative real number. Thus, we must reject the negative root.

Q: Is the Parallelogram Law always applicable here?

Yes! The Parallelogram Law is a fundamental identity in complex numbers that relates the lengths of the diagonals of a parallelogram to the lengths of its sides. Since $\alpha$ and $\beta$ can be viewed as vectors from the origin, the law holds perfectly, and it is the most efficient way to link the center of the circle to the distance between the two fixed points.

Question

For $\alpha, \beta, z \in \mathbb{C}$ and $\lambda > 1$, if $\sqrt{\lambda - 1}$ is the radius of the circle $|z - \alpha|^2 + |z - \beta|^2 = 2\lambda$, then $|\alpha - \beta|$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, flat plane—the complex plane. You have two fixed pegs, $\alpha$ and $\beta$, driven into the ground.

You are holding a string, and you are moving a point $z$ such that the sum of the squares of the distances from $z$ to these two pegs is always a constant, $2\lambda$. This movement traces a perfect circle.

Our mission is to uncover the distance between those two pegs, $|\alpha - \beta|$, using the power of complex algebra.

### The Algebraic Expansion

We begin with the given condition:
$$|z - \alpha|^2 + |z - \beta|^2 = 2\lambda$$

To unlock the secrets hidden in this equation, we use the fundamental identity $|w|^2 = w\bar{w}$. By applying this to each term, we transform the geometric distance into a manageable algebraic expression:
$$(z - \alpha)(\bar{z} - \bar{\alpha}) + (z - \beta)(\bar{z} - \bar{\beta}) = 2\lambda$$

Now, we expand these brackets with care. Multiplying the terms, we get:
$$(z\bar{z} - z\bar{\alpha} - \bar{z}\alpha + |\alpha|^2) + (z\bar{z} - z\bar{\beta} - \bar{z}\beta + |\beta|^2) = 2\lambda$$

### Standardizing the Circle

Let us group the like terms. We have two $z\bar{z}$ terms, which combine to $2z\bar{z}$. We factor out the $z$ and $\bar{z}$ terms to reveal the structure:
$$2z\bar{z} - z(\bar{\alpha} + \bar{\beta}) - \bar{z}(\alpha + \beta) + |\alpha|^2 + |\beta|^2 = 2\lambda$$

To make this look like the standard equation of a circle, $|z|^2 - z\bar{z_0} - \bar{z}z_0 + C = 0$, we divide the entire equation by $2$:
$$|z|^2 - z\left(\frac{\bar{\alpha} + \bar{\beta}}{2}\right) - \bar{z}\left(\frac{\alpha + \beta}{2}\right) + \frac{|\alpha|^2 + |\beta|^2}{2} = \lambda$$

By comparing this to the standard form, we instantly see that the center of our circle is $z_0 = \frac{\alpha + \beta}{2}$, which is the midpoint of the segment joining $\alpha$ and $\beta$.

### The Parallelogram Law Shortcut

Now, we need the radius. The formula for the squared radius of a circle in the complex plane is $r^2 = |z_0|^2 - C$.

Substituting our values, we get:
$$r^2 = \left| \frac{\alpha + \beta}{2} \right|^2 - \left( \frac{|\alpha|^2 + |\beta|^2}{2} - \lambda \right)$$

This is where the magic happens. We invoke the **Parallelogram Law**:
$$|\alpha + \beta|^2 + |\alpha - \beta|^2 = 2(|\alpha|^2 + |\beta|^2)$$

Rearranging this, we can substitute for $|\alpha + \beta|^2$ in our radius equation. When we do this, the terms involving $|\alpha|^2$ and $|\beta|^2$ cancel out with surgical precision, leaving us with:
$$r^2 = \lambda - \frac{|\alpha - \beta|^2}{4}$$

### The Final Revelation

The problem states that the radius is $\sqrt{\lambda - 1}$, so $r^2 = \lambda - 1$. Equating our two expressions for $r^2$, we have:
$$\lambda - 1 = \lambda - \frac{|\alpha - \beta|^2}{4}$$

The $\lambda$ terms vanish, leaving $-1 = -\frac{|\alpha - \beta|^2}{4}$. Multiplying by $-4$, we find $|\alpha - \beta|^2 = 4$.

Taking the positive root, we arrive at the beautiful, simple result:
**$|\alpha - \beta| = 2$**

Animated Solution for Mathematics - Complex Numbers: For $α, β, z \in C$ and $λ > 1$ , if $λ - 1$ is the radius of the circle $∣ z - α ∣^{2} + ∣ z - β ∣^{2} = 2 λ$ , then $∣ α - β ∣$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry in $C$

Expanding the Modulus Squared

Expanding the Terms

Grouping Like Terms

Standardizing the Circle Equation

Identifying the Center $z_{0}$

The Radius Formula

Applying the Parallelogram Law

Simplifying the Radius Expression

Equating with Given Radius

Final Calculation

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Algebraic Expansion

Standardizing the Circle

The Parallelogram Law Shortcut

The Final Revelation

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