Q.If $\left|\dfrac{z-2}{z+2}\right|=\dfrac{\pi}{6}$, then the locus of $z$ is _____.

Concept: Locus of points satisfying a constant ratio of distances to two fixed points — an Apollonius circle.

Let $z=x+iy$. The condition $\left|\dfrac{z-2}{z+2}\right|=\dfrac{\pi}{6}$ means $|z-2| = \dfrac{\pi}{6}|z+2|$, i.e. $6|z-2| = \pi|z+2|$.

Squaring both sides:

Expanding and collecting terms:

Since $\pi^2 \approx 9.87 < 36$, divide through by $(36-\pi^2)$: …

The core idea here is that an equation of the form $\left|\frac{z - a}{z - b}\right| = k$, where $k > 0$ and $k \neq 1$, always represents a circle in the complex plane. This is known as an Apollonius circle — the set of points whose distances to two fixed points are in a constant ratio.

Here, $a = 2$, $b = -2$, and $k = \frac{\pi}{6}$. Since $\pi \approx 3.14$, $\frac{\pi}{6} \approx 0.523$, which is not equal to 1, so the locus is indeed a circle. The centre lies on the line joining the two fixed points — in this case, the real axis.

Let’s derive the equation step by step.

1. Write the condition in algebraic form. Let $z = x + iy$, where $x, y \in \mathbb{R}$. Then:

Cross-multiplying:

2. Square both sides to remove square roots. Squaring is safe because both sides are non-negative:

Recall $|z - z_0|^2 = (x - x_0)^2 + (y - y_0)^2$. So:

3. Expand and simplify.

Bring all terms to one side:

4. Divide through by the common coefficient of $x^2$ and $y^2$. Since $\pi^2 \approx 9.87$, $36 - \pi^2 > 0$, so we can divide:

5. Complete the square in $x$. The equation is of the form $x^2 + y^2 - 2gx + c = 0$, where:

Completing the square:

Here $c = \frac{144 - 4\pi^2}{36 - \pi^2}$. So the radius squared is:

6. Simplify $R^2$. Put everything over a common denominator:

Compute the numerator step by step:

• $(72 + 2\pi^2)^2 = 4(36 + \pi^2)^2 = 4(1296 + 72\pi^2 + \pi^4)$. …