Circles
Complex Locus Intersection — Sum of |z|²
nta_pyq_2026_jan
Grade 11
Question:
Let $S=\left\{z\in\mathbb{C}:\left|\dfrac{z-6i}{z-2i}\right|=1\text{ and }\left|\dfrac{z-8+2i}{z+2i}\right|=\dfrac{3}{5}\right\}$. Then $\displaystyle\sum_{z\in S}|z|^2$ is equal to
Step-by-Step Solution
Key Concept: $|z-6i|=|z-2i|\Rightarrow y=4$ (perpendicular bisector of segment from $2i$ to $6i$). Substitute $z=x+4i$ in the second condition $|z-8+2i|/|z+2i|=3/5$.
$z=17+4i$ and $z=8+4i$. $\sum|z|^2=385$.
Correct Answer: 1