JEE Main 2024 — 27 January, Morning Shift. Previous Year Question.
Problem
If $S = \{z\in\mathbb{C} : |z-i|=|z+i|=|z-1|\}$, then $n(S)$ is:
Key insight. A point equidistant from two fixed points lies on their perpendicular bisector — a standard fact. Since the condition here requires equidistance from three points at once ($i$, $-i$, and $1$), $z$ must lie on the perpendicular bisector of every pair — meaning $z$ is the circumcenter of the triangle formed by these three points, and there’s exactly one such point (unless the three points are collinear).
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Approach
Substitute $z=x+iy$ and translate each pair of equal moduli into an equation. Use $|z-i|=|z+i|$ to get one equation, and $|z-i|=|z-1|$ to get another. Solve the resulting system to find the unique point (or points) satisfying both simultaneously.
Solution
Step 1 — Substitute z = x + iy and set up |z − i| = |z + i|
$$|x+i(y-1)| = |x+i(y+1)|$$
Squaring both sides:
$$x^2+(y-1)^2 = x^2+(y+1)^2$$
$$y^2-2y+1 = y^2+2y+1 \implies -4y=0 \implies y=0$$
Step 2 — Set up |z − i| = |z − 1|
$$x^2+(y-1)^2 = (x-1)^2+y^2$$
Expanding:
$$x^2+y^2-2y+1 = x^2-2x+1+y^2 \implies -2y = -2x \implies y=x$$
Step 3 — Combine both equations
$$y=0 \quad \text{and} \quad y=x \implies x=0,\ y=0$$
Step 4 — Verify z = 0 satisfies all three original conditions
$$|0-i|=1, \qquad |0+i|=1, \qquad |0-1|=1$$
All three are equal, confirming $z=0$ works.
Step 5 — Conclude there is exactly one solution
Since the system $y=0$ and $y=x$ has exactly one solution, $S=\{0\}$, so:
$$n(S)=1$$
Answer
$$1$$
Common mistakes
- Only using one of the two possible equal-modulus pairings. The condition $|z-i|=|z+i|=|z-1|$ is really two independent equations (e.g. $|z-i|=|z+i|$ and $|z-i|=|z-1|$) — using just one leaves an entire line of solutions instead of the single intersection point.
- Assuming the answer is always $1$ for “equidistant from three points.” This only holds when the three given points are not collinear — if they were, no single point could be equidistant from all three (or infinitely many could, if two of the points coincided), giving $n(S)=0$ or a different count entirely.
Practise next
- Find $n(S)$ where $S=\{z\in\mathbb{C} : |z-2i|=|z+2i|=|z-2|\}$, using the same perpendicular-bisector approach.
Show answer
$n(S)=1$. Each equality is a perpendicular bisector, so $S$ is where two of them meet.
$|z-2i|=|z+2i|$ says $z$ is equidistant from $2i$ and $-2i$, which is the real axis: $y=0$.
Then $|z-2i|=|z-2|$ with $z=x$ real gives $\sqrt{x^2+4}=|x-2|$, so $x^2+4=x^2-4x+4$ and $x=0$.
So $z=0$ is the only candidate, and it works: all three distances equal $2$. Hence $S=\{0\}$ — the three points $2i$, $-2i$ and $2$ have a unique circumcentre, as any three non-collinear points do.
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