Complex Numbers – Locus – JEE Main 24 Jan 2026 Shift 1

Solution:

Let $z = x + iy$.

Consider the first condition: $|\frac{z-6i}{z-2i}|=1$
$$|z-6i| = |z-2i|$$
This represents the perpendicular bisector of the segment joining $(0,6)$ and $(0,2)$.
The midpoint is $(0,4)$, and the line is horizontal.
$$y = 4$$

Consider the second condition: $|\frac{z-(8-2i)}{z-(-2i)}|=\frac{3}{5}$
Squaring both sides:
$$|z-(8-2i)|^2 = \frac{9}{25}|z+2i|^2$$
$$25[(x-8)^2 + (y+2)^2] = 9[x^2 + (y+2)^2]$$

Substitute $y=4$ into this equation:
$$25[(x-8)^2 + (4+2)^2] = 9[x^2 + (4+2)^2]$$
$$25[(x-8)^2 + 36] = 9[x^2 + 36]$$
$$25(x^2 – 16x + 64 + 36) = 9(x^2 + 36)$$
$$25(x^2 – 16x + 100) = 9x^2 + 324$$
$$25x^2 – 400x + 2500 = 9x^2 + 324$$
$$16x^2 – 400x + 2176 = 0$$

Divide by 16:
$$x^2 – 25x + 136 = 0$$
$$(x – 17)(x – 8) = 0$$
So, $x = 17$ or $x = 8$.

The points in $S$ are $z_1 = 17 + 4i$ and $z_2 = 8 + 4i$.
We need to find $\sum_{z\in S}|z|^{2} = |z_1|^2 + |z_2|^2$.
$$|z_1|^2 = 17^2 + 4^2 = 289 + 16 = 305$$
$$|z_2|^2 = 8^2 + 4^2 = 64 + 16 = 80$$
$$\text{Sum} = 305 + 80 = 385$$

Ans. (4)