Let C be the circle in the complex plane with centre z0 = 1/2(1 + 3i) and radius r = 1. Let z1 = 1 +

Q: Wait, why can $k$ be negative? Doesn't collinearity just mean they are on the same side?

That is a fantastic question! Collinearity simply means the points lie on the same infinite straight line. It does not restrict them to a specific ray starting from $z_0$. When we say $z_2 - z_0 = k(z_1 - z_0)$, a positive $k$ puts $z_2$ on the same side of $z_0$ as $z_1$, while a negative $k$ puts $z_2$ on the opposite side. Both are perfectly valid collinear positions.

Q: How did you know to use the vector form $z_2 - z_0 = k(z_1 - z_0)$?

In the complex plane, the expression $z_B - z_A$ represents the vector pointing from $A$ to $B$. If three points $z_0, z_1, z_2$ are collinear, the vector from $z_0$ to $z_2$ must be parallel to the vector from $z_0$ to $z_1$. Two vectors are parallel if and only if one is a scalar multiple of the other. This is a powerful tool in complex geometry!

Q: Is there a faster way using inversion geometry?

Yes, absolutely! The condition $|z_1 - z_0||z_2 - z_0| = 1$ is the definition of geometric inversion with respect to a circle centered at $z_0$ with radius $1$. In this context, $z_1$ and $z_2$ are inverse points. However, for JEE Advanced, the algebraic vector approach is often safer and less prone to conceptual errors regarding the orientation of the points.

Question

Let $C$ be the circle in the complex plane with centre $z_0 = \frac{1}{2}(1 + 3i)$ and radius $r = 1$. Let $z_1 = 1 + i$ and the complex number $z_2$ be outside circle $C$ such that $|z_1 - z_0||z_2 - z_0| = 1$. If $z_0, z_1, z_2$ are collinear, then the smaller value of $|z_2|^2$ is equal to

Options:

(A) $5/2$

(B) $7/2$

(C) $13/2$

(D) $3/2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler, to the beautiful landscape of the complex plane. Today, we are not just solving an equation; we are mapping a relationship between points in a two-dimensional space.

We have a circle $C$ centered at $z_0 = \frac{1}{2}(1 + 3i)$ with a radius $r = 1$. We are also given a point $z_1 = 1 + i$.

Our mission is to find a point $z_2$ that is outside this circle, collinear with $z_0$ and $z_1$, and satisfies the distance product: $|z_1 - z_0||z_2 - z_0| = 1$.

### The Distance Bridge

Before we dive into the algebra, let us ground ourselves in the geometry. We need to know how far $z_1$ is from the center $z_0$.

We calculate the magnitude of the difference:
$$|z_1 - z_0| = |(1 + i) - (0.5 + 1.5i)| = |0.5 - 0.5i|$$

To find the magnitude, we take the square root of the sum of the squares of the real and imaginary parts:
$$|z_1 - z_0| = \sqrt{(0.5)^2 + (-0.5)^2} = \sqrt{0.25 + 0.25} = \sqrt{0.5} = \frac{1}{\sqrt{2}}$$

Now, look at the magic of the product relation given in the problem: $|z_1 - z_0| \cdot |z_2 - z_0| = 1$. Substituting our known distance, we get:
$$\frac{1}{\sqrt{2}} \cdot |z_2 - z_0| = 1 \Rightarrow |z_2 - z_0| = \sqrt{2}$$

### The Collinearity Constraint

Now, we invoke the condition of collinearity. If $z_0, z_1,$ and $z_2$ are collinear, they lie on the same line.

In the language of complex numbers, this means the vector from $z_0$ to $z_2$ is a scaled version of the vector from $z_0$ to $z_1$. We write this as $z_2 - z_0 = k(z_1 - z_0)$, where $k$ is a real scalar.

Taking the magnitude of both sides, we get $|z_2 - z_0| = |k| |z_1 - z_0|$. Plugging in our known values:
$$\sqrt{2} = |k| \cdot \frac{1}{\sqrt{2}} \Rightarrow |k| = 2$$

This is the crucial moment where the problem splits into two paths: $k = 2$ or $k = -2$.

### The Two Worlds

Let us explore these two possibilities. If $k = 2$, then $z_2 - z_0 = 2(z_1 - z_0)$, which means $z_2 = 2z_1 - z_0$.

Substituting the values:
$$z_2 = 2(1 + i) - (0.5 + 1.5i) = 2 + 2i - 0.5 - 1.5i = 1.5 + 0.5i$$

For this candidate, the squared magnitude is:
$$|z_2|^2 = (1.5)^2 + (0.5)^2 = 2.25 + 0.25 = 2.5 = \frac{5}{2}$$

Now, consider the second case where $k = -2$. Then $z_2 - z_0 = -2(z_1 - z_0)$, which leads to $z_2 = z_0 - 2(z_1 - z_0) = 3z_0 - 2z_1$.

Substituting the values:
$$z_2 = 3(0.5 + 1.5i) - 2(1 + i) = 1.5 + 4.5i - 2 - 2i = -0.5 + 2.5i$$

For this second candidate, the squared magnitude is:
$$|z_2|^2 = (-0.5)^2 + (2.5)^2 = 0.25 + 6.25 = 6.5 = \frac{13}{2}$$

### The Final Reflection

We have arrived at two possible values for $|z_2|^2$: $\frac{5}{2}$ and $\frac{13}{2}$. The problem asks for the smaller value.

Comparing the two, it is clear that **$\frac{5}{2}$** is the smaller one. We have navigated the complex plane, respected the geometric constraints, and arrived at the solution.

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Visualized Solution

Plotting $z_{0}$ and $z_{1}$

Visualizing Circle $C$

Distance Between $z_{1}$ and $z_{0}$

Calculating $∣ z_{1} - z_{0} ∣$

Using the Product Relation

The Collinearity Condition

Finding the Scalar $k$

Case 1: $k = 2$

Calculating $∣ z_{2} ∣^{2}$ for Case 1

Case 2: $k = - 2$

Calculating $∣ z_{2} ∣^{2}$ for Case 2

Comparing and Final Result

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Distance Bridge

The Collinearity Constraint

The Two Worlds

The Final Reflection

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Visualized Solution

Plotting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z0​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z1​

Visualizing Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Distance Between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z0​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣z1​−z0​∣