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NCERT Exemplar · Q45

Q.If ∣z−2z+2∣=π6\left|\dfrac{z-2}{z+2}\right|=\dfrac{\pi}{6}, then the locus of zz is _____.

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The given condition ∣z−2z+2∣=π6\left|\frac{z-2}{z+2}\right| = \frac{\pi}{6} describes a circle (Apollonius circle) in the complex plane. The locus is a circle with centre on the real axis, specifically at (2(36+π2)36−π2,0)\left(\frac{2(36+\pi^2)}{36-\pi^2}, 0\right) and radius 24π∣π2−36∣\frac{24\pi}{|\pi^2 - 36|}.

The core idea here is that an equation of the form ∣z−az−b∣=k\left|\frac{z - a}{z - b}\right| = k, where k>0k > 0 and k≠1k \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=2a = 2, b=−2b = -2, and k=π6k = \frac{\pi}{6}. Since π≈3.14\pi \approx 3.14, π6≈0.523\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+iyz = x + iy, where x,y∈Rx, y \in \mathbb{R}. Then:

∣z−2z+2∣=π6⇒∣z−2∣∣z+2∣=π6.\left|\frac{z-2}{z+2}\right| = \frac{\pi}{6} \quad\Rightarrow\quad \frac{|z-2|}{|z+2|} = \frac{\pi}{6}.

Cross-multiplying:

6∣z−2∣=π∣z+2∣.6|z-2| = \pi |z+2|.

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

36∣z−2∣2=π2∣z+2∣2.36 |z-2|^2 = \pi^2 |z+2|^2.

Recall ∣z−z0∣2=(x−x0)2+(y−y0)2|z - z_0|^2 = (x - x_0)^2 + (y - y_0)^2. So:

36[(x−2)2+y2]=π2[(x+2)2+y2].36\left[(x-2)^2 + y^2\right] = \pi^2\left[(x+2)^2 + y^2\right].

  1. Expand and simplify.

36(x2−4x+4+y2)=π2(x2+4x+4+y2).36(x^2 - 4x + 4 + y^2) = \pi^2(x^2 + 4x + 4 + y^2).

36x2−144x+144+36y2=π2x2+4π2x+4π2+π2y2.36x^2 - 144x + 144 + 36y^2 = \pi^2 x^2 + 4\pi^2 x + 4\pi^2 + \pi^2 y^2.

Bring all terms to one side:

(36−π2)x2+(36−π2)y2−(144+4π2)x+(144−4π2)=0.(36 - \pi^2)x^2 + (36 - \pi^2)y^2 - (144 + 4\pi^2)x + (144 - 4\pi^2) = 0.

  1. Divide through by the common coefficient of x2x^2 and y2y^2. Since π2≈9.87\pi^2 \approx 9.87, 36−π2>036 - \pi^2 > 0, so we can divide:

x2+y2−144+4π236−π2 x+144−4π236−π2=0.x^2 + y^2 - \frac{144 + 4\pi^2}{36 - \pi^2}\,x + \frac{144 - 4\pi^2}{36 - \pi^2} = 0.

  1. Complete the square in xx. The equation is of the form x2+y2−2gx+c=0x^2 + y^2 - 2gx + c = 0, where:

2g=144+4π236−π2⇒g=72+2π236−π2.2g = \frac{144 + 4\pi^2}{36 - \pi^2} \quad\Rightarrow\quad g = \frac{72 + 2\pi^2}{36 - \pi^2}.

Completing the square:

(x−g)2+y2=g2−c.(x - g)^2 + y^2 = g^2 - c.

Here c=144−4π236−π2c = \frac{144 - 4\pi^2}{36 - \pi^2}. So the radius squared is:

R2=g2−c=(72+2π236−π2)2−144−4π236−π2.R^2 = g^2 - c = \left(\frac{72 + 2\pi^2}{36 - \pi^2}\right)^2 - \frac{144 - 4\pi^2}{36 - \pi^2}.

  1. Simplify R2R^2. Put everything over a common denominator:

R2=(72+2π2)2−(144−4π2)(36−π2)(36−π2)2.R^2 = \frac{(72 + 2\pi^2)^2 - (144 - 4\pi^2)(36 - \pi^2)}{(36 - \pi^2)^2}.

Compute the numerator step by step:

  • (72+2π2)2=4(36+π2)2=4(1296+72π2+π4)(72 + 2\pi^2)^2 = 4(36 + \pi^2)^2 = 4(1296 + 72\pi^2 + \pi^4). …
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