If the center and radius of the circle |(z-2)/(z-3)| = 2 are respectively (α, β) and γ, then 3(α + β

Q: Wait, why do we square both sides? Can't we just solve it with the modulus directly?

You absolutely could, but it would be like trying to cut a diamond with a butter knife! The modulus $|z-z_0|$ involves a square root: $\sqrt{(x-x_0)^2 + (y-y_0)^2}$. Squaring both sides immediately removes that square root, turning a messy radical equation into a clean, manageable polynomial equation. It is the most efficient path to the solution.

Q: How do I know for sure that this locus is a circle?

That is a fantastic question! In the Cartesian plane, any equation of the form $Ax^2 + Ay^2 + Dx + Ey + F = 0$ (where the coefficients of $x^2$ and $y^2$ are identical) represents a circle. When we expanded our equation, we ended up with $3x^2 + 3y^2 - 20x + 32 = 0$. Since the coefficients of $x^2$ and $y^2$ are both $3$, the geometry is guaranteed to be circular.

Q: What if the ratio was $1$ instead of $2$? Would it still be a circle?

If the ratio $k$ were $1$, the equation would become $|z-2| = |z-3|$. Geometrically, this means the distance from $z$ to $2$ equals the distance from $z$ to $3$. The locus of points equidistant from two fixed points is the perpendicular bisector of the segment joining them—a straight line, not a circle! So, the circle only exists when $k \neq 1$.

Question

If the center and radius of the circle $|\frac{z-2}{z-3}| = 2$ are respectively $(\alpha, \beta)$ and $\gamma$, then $3(\alpha + \beta + \gamma)$ is equal to

Options:

(A) 11

(B) 9

(C) 10

(D) 12

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 a hidden geometric truth.

We are given the complex equation:

$$ \left| \frac{z-2}{z-3} \right| = 2 $$

At first glance, it looks like a simple algebraic expression, but beneath the surface lies the elegant geometry of the **Apollonius Circle**. Imagine you are standing on the complex plane with two fixed beacons at $2$ and $3$ on the real axis. The equation dictates that the point $z$ moves such that its distance from $2$ is always exactly twice its distance from $3$.

### Phase 1: The Algebraic Transformation

We begin by cross-multiplying to clear the fraction, yielding $|z-2| = 2|z-3|$. To eliminate the radicals inherent in the modulus, we square both sides:

$$ |z-2|^2 = 4|z-3|^2 $$

This is a crucial step. By substituting $z = x + iy$, where $x$ and $y$ are the real and imaginary parts, we transform the complex modulus into a standard Cartesian form:

$$ |(x-2) + iy|^2 = 4|(x-3) + iy|^2 $$

Using the property $|a+ib|^2 = a^2 + b^2$, we obtain the expanded equation:

$$ (x-2)^2 + y^2 = 4((x-3)^2 + y^2) $$

### Phase 2: The Emergence of the Circle

Next, we expand the binomials. The left side becomes $x^2 - 4x + 4 + y^2$, and the right side becomes $4(x^2 - 6x + 9 + y^2)$, which simplifies to $4x^2 - 24x + 36 + 4y^2$.

Bringing all terms to one side, we get:

$$ 3x^2 + 3y^2 - 20x + 32 = 0 $$

The fact that the coefficients of $x^2$ and $y^2$ are identical confirms this is a circle. To find the center $(\alpha, \beta)$ and radius $\gamma$, we normalize the coefficients by dividing by $3$:

$$ x^2 + y^2 - \frac{20}{3}x + \frac{32}{3} = 0 $$

### Phase 3: The Final Calculation

Comparing this to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$, we identify $2g = -\frac{20}{3}$ and $2f = 0$. Thus, $g = -\frac{10}{3}$ and $f = 0$.

The center of the circle $(\alpha, \beta)$ is $(-g, -f)$, which gives us:

$$ \alpha = \frac{10}{3}, \quad \beta = 0 $$

To find the radius $\gamma$, we use the formula $\gamma = \sqrt{g^2 + f^2 - c}$:

$$ \gamma = \sqrt{\left(-\frac{10}{3}\right)^2 + 0^2 - \frac{32}{3}} = \sqrt{\frac{100}{9} - \frac{96}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3} $$

### Conclusion: The Beauty of the Result

We have determined the parameters: $\alpha = \frac{10}{3}$, $\beta = 0$, and $\gamma = \frac{2}{3}$. The problem asks for the value of $3(\alpha + \beta + \gamma)$.

Substituting our values, we calculate:

$$ 3\left(\frac{10}{3} + 0 + \frac{2}{3}\right) = 3\left(\frac{12}{3}\right) = 12 $$

The complexity of the initial equation has collapsed into a beautiful, simple integer. The final answer is **12**.

Animated Solution for Mathematics - Complex Numbers: If the center and radius of the circle $∣ \frac{z - 2}{z - 3} ∣ = 2$ are respectively $(α, β)$ and $γ$ , then $3 (α + β + γ)$ is equal to

Select Answer:

Visualized Solution

The Complex Plane and Fixed Points

Understanding the Locus Equation

Squaring to Remove Modulus

Substituting $z = x + i y$

Applying the Modulus Formula

Expanding the Squares

Distributing the Constant

Rearranging the Equation

Standard Form of the Circle

Identifying the Center $(α, β)$

Calculating the Radius $γ$

Final Evaluation

Summary and Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Conclusion

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