Let |(z_bar - i)/(2z_bar + i)| = 1/3, z ∈ C be the equation of a circle with center at C. If the are

Q: Why do we square both sides instead of solving for z directly?

Great question! While you could try to manipulate the complex numbers directly, the goal here is to find the center of a circle, which is a geometric entity in the Cartesian plane. By squaring both sides, we effectively map the complex equation into a standard algebraic equation of a circle in the $xy$-plane, which makes identifying the center $C$ and calculating the area of the triangle much more intuitive.

Q: Wait, why is the center $C$ at $(0, -11/5)$? I thought the general equation was $x^2 + y^2 + 2gx + 2fy + c = 0$.

You are absolutely right about the general form! After we simplified our equation to $x^2 + y^2 + \frac{22}{5}y + \frac{8}{5} = 0$, we compared it to the general form. Since there is no $x$ term, $2g = 0$, so $g=0$. For the $y$ term, $2f = \frac{22}{5}$, which gives $f = \frac{11}{5}$. The center is $(-g, -f)$, which leads us directly to $(0, -\frac{11}{5})$. It is all about careful comparison!

Q: Does the sign of $\alpha$ matter for the final answer?

Not at all! The problem asks for $\alpha^2$. Whether $\alpha$ is positive or negative, its square will be positive. In our calculation, we found $|\alpha| = 10$, so $\alpha$ could be $10$ or $-10$. Either way, $\alpha^2 = 100$. The geometry of the triangle ensures the area is positive, but the final algebraic step is robust against the sign of $\alpha$.

Question

Let $|\frac{\overline{z}-i}{2\overline{z}+i}|=\frac{1}{3}, z\in C$ be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the points $(0,0), C$ and $(\alpha,0)$ is 11 square units, then $\alpha^{2}$ equals:

Options:

(A) 50

(B) 100

(C) $\frac{81}{25}$

(D) $\frac{121}{25}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an equation; we are uncovering the hidden geometry within the complex plane.

The problem presents us with a seemingly daunting equation:

$$ \left| \frac{\overline{z}-i}{2\overline{z}+i} ight| = \frac{1}{3} $$

It looks like a mess of conjugates and fractions, but beneath this algebraic exterior lies a beautiful, symmetric circle waiting to be revealed.

### Phase 1: The Modulus Transformation

Our first task is to peel back the layers. We define our complex number as $z = x + iy$, which immediately gives us its conjugate $\overline{z} = x - iy$.

Substituting this into our equation, we get:

$$ \left| \frac{(x-iy)-i}{2(x-iy)+i} ight| = \frac{1}{3} $$

Now, we group the real and imaginary parts. The numerator becomes $x - i(y+1)$, and the denominator becomes $2x + i(1-2y)$.

This is where we invoke the powerful property of the modulus: $|\frac{z_1}{z_2}| = \frac{|z_1|}{|z_2|}$. By applying this, we can rewrite our equation as:

$$ \frac{|x - i(y+1)|}{|2x + i(1-2y)|} = \frac{1}{3} $$

Cross-multiplying gives us $3|x - i(y+1)| = |2x + i(1-2y)|$. We have successfully tamed the fraction.

### Phase 2: The Geometric Reveal

Now, we face the modulus. To eliminate it, we square both sides, transforming the complex modulus into a sum of squares:

$$ 9(x^2 + (y+1)^2) = (2x)^2 + (1-2y)^2 $$

Expanding the left side, we get $9(x^2 + y^2 + 2y + 1)$. On the right, we have $4x^2 + (1 - 4y + 4y^2)$.

Distributing the $9$ and bringing everything to one side, we arrive at:

$$ 5x^2 + 5y^2 + 22y + 8 = 0 $$

Dividing by $5$, we get the standard form:

$$ x^2 + y^2 + \frac{22}{5}y + \frac{8}{5} = 0 $$

Comparing this to the general circle equation $x^2 + y^2 + 2gx + 2fy + c = 0$, we identify the center $C$ as $(0, -\frac{11}{5})$. The circle is centered on the negative $y$-axis.

### Phase 3: The Final Geometry

We are now in the home stretch. We have a triangle with vertices at the origin $O(0,0)$, the center $C(0, -\frac{11}{5})$, and a point $A(\alpha, 0)$.

Because $C$ lies on the $y$-axis and $A$ lies on the $x$-axis, the angle at the origin is $90^\circ$. This is a right-angled triangle.

The area is simply $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. The base is $|\alpha|$ and the height is $|-\frac{11}{5}| = \frac{11}{5}$.

Setting the area to $11$, we have:

$$ \frac{1}{2} 	imes |\alpha| 	imes \frac{11}{5} = 11 $$

Solving this, we find $|\alpha| = 10$. Squaring this gives us **$\alpha^2 = 100$**.

We have arrived at our destination. The complexity of the start has dissolved into the elegance of the result.

Animated Solution for Mathematics - Complex Numbers: Let $∣ \frac{z - i}{2 z + i} ∣ = \frac{1}{3}, z \in C$ be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the points $(0, 0), C$ and $(α, 0)$ is 11 square units, then $α^{2}$ equals:

Select Answer:

Visualized Solution

Define $z$ and $\overline{z}$

Group Real and Imaginary Parts

Apply Modulus Property

Square Both Sides

Expand the Algebraic Terms

Simplify to Circle Form

Standard Circle Equation

Identify the Center $C$

Visualize the Triangle Vertices

Form the Right-Angled Triangle

Apply Area Formula

Solve for $∣ α ∣$

Calculate $α^{2}$

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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