Find the centre and radius of circle given by |(z-α)/(z-β)| = k, k ≠ 1 where z = x + iy, α = α1 + iα

Q: Wait, why do we reject $k=1$ in the problem statement?

That is a brilliant observation! If $k=1$, the equation becomes $|z-\alpha| = |z-\beta|$. Geometrically, this is the locus of points equidistant from $\alpha$ and $\beta$, which is the perpendicular bisector of the segment joining them. A straight line is essentially a circle with an infinite radius. Our algebraic derivation involves dividing by $(1-k^2)$, which would be undefined if $k=1$. So, the condition $k \neq 1$ ensures we are dealing with a finite circle.

Q: Why can't we just substitute $z = x+iy$ and solve it like a coordinate geometry problem?

You *could*, but it would be an algebraic nightmare involving square roots and massive expansions. In JEE Advanced, we value elegance and efficiency. Using complex conjugates keeps the structure clean and allows us to see the circle's properties directly without getting lost in a sea of $x$ and $y$ variables.

Q: How do I know the center is $-a$ and the radius is $\sqrt{|a|^2 - b}$?

The standard form of a circle in the complex plane is $z\bar{z} + \bar{a}z + a\bar{z} + b = 0$. By comparing the coefficients of our derived equation to this standard form, we identify the center as $-a$. The radius formula is derived by completing the square in the complex plane, resulting in $\sqrt{|a|^2 - b}$. It is a standard result, but deriving it once gives you immense confidence!

Question

Find the centre and radius of circle given by $|\frac{z-\alpha}{z-\beta}| = k, k 
eq 1$ where $z = x + iy, \alpha = \alpha_1 + i\alpha_2, \beta = \beta_1 + i\beta_2$.

Sigma Percentile · Accepted Answer

### The Geometry of Ratios: Unveiling the Apollonius Circle

Welcome, future engineers. Today, we are not just solving an equation; we are embarking on a journey into the heart of complex geometry.

When you look at the equation $\frac{|z-\alpha|}{|z-\beta|} = k$, what do you see? Do you see a terrifying mess of moduli and variables, or do you see a beautiful, hidden symmetry? Let us peel back the layers together.

### Phase 1: The Geometric Intuition

Imagine you are standing on a vast, flat plane with two fixed markers, $\alpha$ and $\beta$. Now, imagine a point $z$ that moves in such a way that its distance from $\alpha$ is always a fixed multiple $k$ of its distance from $\beta$.

If $k=1$, you are simply walking along the perpendicular bisector of the segment connecting $\alpha$ and $\beta$. But what happens when $k 
eq 1$?

This is the **Apollonius Circle**. It is a classic, elegant locus where the condition $\frac{|z-\alpha|}{|z-\beta|} = k$ forces $z$ to trace a perfect circle. Our goal is to find the center and radius of this circle without losing our sanity in the algebra.

### Phase 2: The Algebraic Bridge

To tame this beast, we must avoid the trap of Cartesian coordinates. Do not substitute $z = x+iy$ yet, as that path is fraught with square roots and tedious expansions.

Instead, let us use the most powerful tool in our complex number toolkit: the property $|w|^2 = w\bar{w}$. First, let us clear the fraction and square both sides to eliminate the modulus:

$$|z-\alpha|^2 = k^2 |z-\beta|^2$$

Now, we apply our magic property. We replace the squared modulus with the product of the complex number and its conjugate:

$$(z-\alpha)(\bar{z}-\bar{\alpha}) = k^2 (z-\beta)(\bar{z}-\bar{\beta})$$

### Phase 3: The Expansion

This is where many students stumble, but you will not. Expand both sides carefully, remembering that $\alpha\bar{\alpha} = |\alpha|^2$ and $\beta\bar{\beta} = |\beta|^2$.

On the left side, we have:

$$z\bar{z} - \bar{\alpha}z - \alpha\bar{z} + |\alpha|^2$$

On the right side, we have:

$$k^2(z\bar{z} - \bar{\beta}z - \beta\bar{z} + |\beta|^2)$$

Now, bring everything to one side to group the $z\bar{z}$ terms, the $z$ terms, and the $\bar{z}$ terms:

$$(1-k^2)z\bar{z} - (\bar{\alpha}-k^2\bar{\beta})z - (\alpha-k^2\beta)\bar{z} + |\alpha|^2 - k^2|\beta|^2 = 0$$

### Phase 4: The Elegant Conclusion

To find the center and radius, we need the coefficient of $z\bar{z}$ to be unity. We divide the entire equation by $(1-k^2)$, which is valid since the problem explicitly states $k 
eq 1$:

$$z\bar{z} - \frac{\bar{\alpha}-k^2\bar{\beta}}{1-k^2}z - \frac{\alpha-k^2\beta}{1-k^2}\bar{z} + \frac{|\alpha|^2 - k^2|\beta|^2}{1-k^2} = 0$$

Comparing this to the standard form $z\bar{z} + \bar{a}z + a\bar{z} + b = 0$, we identify our center as $-a$. Looking at the coefficient of $\bar{z}$, we see that $a = -\frac{\alpha-k^2\beta}{1-k^2}$.

Therefore, the **Centre** is:

$$	ext{Centre} = \frac{\alpha - k^2\beta}{1 - k^2}$$

Finally, for the radius, we use the formula $R = \sqrt{|a|^2 - b}$. After substituting our values and simplifying the expression, the terms cancel out with satisfying precision, leaving us with the **Radius**:

$$	ext{Radius} = \frac{k}{|1 - k^2|} |\alpha - \beta|$$

### Final Thoughts

Look at that result. It is symmetric, clean, and powerful. You have successfully navigated the Apollonius Circle.

You didn't just calculate a center and radius; you uncovered the geometric structure of a complex ratio. Keep this mindset—always look for the elegant path, trust your algebraic tools, and never fear the complexity. You are ready for the next challenge.

Animated Solution for Mathematics - Complex Numbers: Find the centre and radius of circle given by $∣ \frac{z - α}{z - β} ∣ = k, k \neq = 1$ where $z = x + i y, α = α_{1} + i α_{2}, β = β_{1} + i β_{2}$ .

Visualized Solution

Visualizing the Locus

The Apollonius Circle

Squaring the Equation

Using Modulus Property

Expanding the Terms

Grouping the Variables

Standard Circle Equation

Identifying the Centre

Calculating the Radius

The Final Radius

The Sigma Insight: Geometrical Applications of Complex Numbers

The Geometry of Ratios

Phase 1

Phase 2

Phase 3

Phase 4

Final Thoughts

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