JEE Mathematics 2026 Question with Solution

Given:

$$\left|\frac{z-6i}{z-2i}\right|=1$$

and

$$\left|\frac{z-8+2i}{z+2i}\right|=\frac{3}{5}$$

Find: $\sum_{z\in S}|z|^2$

From the first condition,

$$\left|\frac{z-6i}{z-2i}\right|=1 \Rightarrow |z-6i|=|z-2i|$$

This represents the perpendicular bisector of the line joining $6i$ and $2i$. Hence,

$$\operatorname{Im}(z)=4$$

Let

$$z=x+iy$$

so that

$$y=4$$

From the second condition,

$$\left|\frac{z-(8-2i)}{z+2i}\right|=\frac{3}{5}$$

so

$$\frac{|z-(8-2i)|}{|z+2i|}=\frac{3}{5}$$

Squaring both sides,

$$25|z-(8-2i)|^2=9|z+2i|^2$$

Substituting $z=x+iy$ and $y=4$,

$$25\left[(x-8)^2+(4+2)^2\right]=9\left[x^2+(4+2)^2\right]$$

That is,

$$25\left[(x-8)^2+36\right]=9\left(x^2+36\right)$$

Solving,

$$x=2 \text{ or } x=14$$

Thus,

$$z_1=2+4i, \quad z_2=14+4i$$

Now,

$$|z_1|^2=2^2+4^2=20$$

and

$$|z_2|^2=14^2+4^2=212$$

Therefore,

$$\sum |z|^2=20+212=398$$

So, the correct option is A.

Given: The first modulus equation gives points equidistant from $6i$ and $2i$.

Find: The elements of $S$ and then compute $\sum_{z\in S}|z|^2$.

The locus

$$|z-6i|=|z-2i|$$

is the perpendicular bisector of the segment joining the points $0+6i$ and $0+2i$ in the complex plane. Therefore the imaginary part is fixed as

$$\operatorname{Im}(z)=4$$

So every point in $S$ lies on the line

$$y=4$$

Using this in the second condition gives two intersection points, namely

$$z_1=2+4i, \quad z_2=14+4i$$

Then

$$|z_1|^2=20, \quad |z_2|^2=212$$

Hence,

$$\sum_{z\in S}|z|^2=398$$

Therefore, the correct option is A.