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Module 12 · L13 of 16 ~40 min ⚡ +90 XP available

Loci — Circles

A locus is the set of points satisfying a geometric condition. In the complex plane, conditions on $|z - z_0|$ (a distance) trace out circles. This lesson shows how $|z - z_0| = r$ gives a circle of radius $r$ about $z_0$, how the Apollonius condition $|z-a|/|z-b| = k$ ($k \neq 1$) hides a circle inside a ratio, and how to switch between complex and Cartesian forms with confidence.

Today's hook — Sketch the set of points $z$ in the complex plane satisfying $|z - (2 + i)| = 3$. Predict before solving: what is the centre, what is the radius, and where does the circle cross the real axis? Verify by converting to Cartesian: $(x-2)^2 + (y-1)^2 = 9$.

0/5QUESTS

Recall — your gut answer first

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For $z = x + iy$ and $z_0 = a + ib$, what does $|z - z_0|$ measure geometrically? Before checking — write the expression $|z - z_0|^2$ purely in terms of $x, y, a, b$, then say in one sentence why $|z - z_0| = r$ traces a circle.

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The two moves for a complex locus

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Every locus problem in the complex plane rewards two habits: read the condition geometrically first (distance? ratio? angle?), then substitute $z = x + iy$ and simplify only if you need a Cartesian equation. Trying to algebra-brute-force a condition that screams "circle" is the single biggest waste of time in MEX-N1.

The geometry-then-algebra reading: (1) name the shape from the form of the condition, (2) read off centre/radius (or other parameters), (3) only convert to $x, y$ if the question asks.

Circle: $|z - z_0| = r$ · Apollonius: $|z-a|/|z-b| = k$ · Unit circle: $|z| = 1$

$|z - z_0| = r \;\Leftrightarrow\; (x-a)^2 + (y-b)^2 = r^2$

Distance reads circle

If $|z - z_0|$ equals a constant $r > 0$, the locus is a circle of radius $r$ centred at $z_0$. No algebra needed to identify it.

Ratio of distances is also a circle

$|z - a|/|z - b| = k$ with $k > 0$, $k \neq 1$ defines an Apollonius circle. When $k = 1$ the locus collapses to the perpendicular bisector of $ab$ — a straight line.

Unit circle and rotation

$|z| = 1$ is the unit circle. Multiplying any $z$ on it by $e^{i\theta}$ rotates it through $\theta$ around the origin without changing $|z|$ — a key fact for later modulus-argument work.

What you'll master

Know

Key facts

$|z - z_0| = r$ defines a circle of radius $r$ centred at $z_0$
$|z| = 1$ is the unit circle centred at the origin
$|z - a|/|z - b| = k$ ($k \neq 1$) defines an Apollonius circle
Cartesian form: $(x-a)^2 + (y-b)^2 = r^2$ with $z_0 = a + ib$

Understand

Concepts

Why $|z - z_0|$ is the distance from $z$ to $z_0$ in the Argand plane
Why a fixed ratio of two distances ($k \neq 1$) gives a circle, not a line
Why $k = 1$ degenerates the Apollonius locus to a straight line (perpendicular bisector)

Can do

Skills

Convert between $|z - z_0| = r$ and Cartesian form $(x-a)^2 + (y-b)^2 = r^2$
Identify centre and radius of an Apollonius circle from $|z - a|/|z - b| = k$
Sketch loci on an Argand diagram and mark intercepts with the real and imaginary axes

Key terms

LocusThe set of all points satisfying a given geometric condition. In the complex plane, conditions are written in terms of $z$.

Modulus $|z|$The distance from $z = x + iy$ to the origin. $|z| = \sqrt{x^2 + y^2}$. More generally, $|z - z_0|$ is the distance from $z$ to $z_0$.

Circle $|z - z_0| = r$The locus of points at fixed distance $r$ from the fixed point $z_0$. Centre $z_0 = a + ib$, radius $r > 0$.

Apollonius circleLocus of points $z$ such that the ratio of distances to two fixed points $a$ and $b$ is constant: $|z - a|/|z - b| = k$, $k > 0$, $k \neq 1$.

Unit circle$|z| = 1$ — the circle of radius 1 centred at the origin. Every point on it has the form $e^{i\theta} = \cos\theta + i\sin\theta$.

Argand diagramCartesian plane used to plot complex numbers, with the real part as the $x$-coordinate and the imaginary part as the $y$-coordinate.

MEX-N1NESA outcome (Complex Numbers I): represent loci in the complex plane, including circles defined by modulus conditions.

Distance form: $|z - z_0| = r$

core concept

For any fixed complex number $z_0 = a + ib$ and any positive real $r$, the equation $|z - z_0| = r$ describes all complex numbers $z$ whose distance to $z_0$ equals $r$. That is exactly the definition of a circle.

Centre $z_0 = a + ib$ — the fixed point in the modulus.
Radius $r$ — the constant on the right.
Cartesian form: square both sides and expand. $|z - z_0|^2 = (x-a)^2 + (y-b)^2 = r^2$.

Worked through the hook: $|z - (2 + i)| = 3$. Centre $z_0 = 2 + i$, so $(a, b) = (2, 1)$. Radius $r = 3$.

Cartesian: $(x - 2)^2 + (y - 1)^2 = 9$.
Real-axis intercepts: set $y = 0$: $(x - 2)^2 + 1 = 9$, so $x = 2 \pm 2\sqrt{2}$.
Imaginary-axis intercepts: set $x = 0$: $4 + (y - 1)^2 = 9$, so $y = 1 \pm \sqrt{5}$.

Connecting to vectors. Geometrically, $z - z_0$ is the vector from $z_0$ to $z$. Asking for $|z - z_0| = r$ is asking that vector to have fixed length $r$ — its tip sweeps out a circle.

$|z - z_0| = r \Leftrightarrow$ circle, centre $z_0$, radius $r$ · Cartesian: $(x - a)^2 + (y - b)^2 = r^2$ where $z_0 = a + ib$ · $|z| = r$ is the special case $z_0 = 0$ (circle centred at origin) · To find axis intercepts: set $y = 0$ or $x = 0$ in the Cartesian form

Pause — copy $|z-z_0|=r \Leftrightarrow (x-a)^2+(y-b)^2=r^2$, the special case $|z|=r$ (centred at origin), and the axis-intercept method (set $y=0$ or $x=0$) into your book.

Quick check: The locus $|z - (3 - 2i)| = 5$ in the Argand plane is a circle. What is its Cartesian equation?

Apollonius circles: $|z - a|/|z - b| = k$

core concept

We just saw that $|z-z_0| = r$ gives a circle with centre $z_0$ and radius $r$, equivalent to $(x-a)^2+(y-b)^2 = r^2$ in Cartesian form. That raises a question: what if the condition is a ratio of distances $|z-a|/|z-b| = k$? This card answers it → for $k \neq 1$ this is an Apollonius circle, found by squaring, expanding, and completing the square.

Given two distinct fixed points $a$ and $b$ in the complex plane and a positive constant $k$, the locus $|z - a|/|z - b| = k$ is:

The perpendicular bisector of the segment $ab$ when $k = 1$ (a straight line).
A circle — the Apollonius circle — when $k > 0$ and $k \neq 1$.

Method. Square both sides to clear the modulus: $|z - a|^2 = k^2 |z - b|^2$. With $z = x + iy$, expand both sides, collect $x^2 + y^2$ terms (their coefficients are $1$ and $k^2$, so they do not cancel when $k \neq 1$), then divide through to get a Cartesian circle equation. Complete the square to read off centre and radius.

$$|z - a|^2 = k^2 \, |z - b|^2 \;\Rightarrow\; (1 - k^2)(x^2 + y^2) + \cdots = 0$$

Common mistake. Students often expect a line and stop after one round of squaring. With $k \neq 1$, the $x^2 + y^2$ terms do not cancel — so the equation is a circle. Always complete the square to confirm centre and radius rather than guessing them.

$|z - a|/|z - b| = 1 \Rightarrow$ perpendicular bisector of $ab$ (line) · $|z - a|/|z - b| = k$, $k \neq 1 \Rightarrow$ Apollonius circle · Method: square, expand with $z = x + iy$, divide by $1 - k^2$, complete the square · Centre and radius pop out from the completed square — do not guess

Pause — copy the Apollonius locus: $|z-a|/|z-b|=1$ gives a perpendicular bisector (line) and $k \neq 1$ gives a circle; the method is to square, expand, divide by $1-k^2$, complete the square into your book.

Did you get this? True or false: the locus $|z - 2|/|z + 2| = 1$ in the Argand plane is a circle.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · CIRCLE FROM MODULUS

Sketch the locus of $z$ in the Argand plane satisfying $|z - 1 + 2i| = 4$. State the centre, radius, and the points where the circle crosses the real axis.

Rewrite as $|z - (1 - 2i)| = 4$. So $z_0 = 1 - 2i$, $r = 4$. Centre is $(1, -2)$, radius $4$.

Always pull the modulus into the form $|z - z_0|$ before reading off the centre. The sign of the constant inside flips when you do this.

PROBLEM 2 · APOLLONIUS CIRCLE

Find the Cartesian equation of the locus of $z$ such that $|z - 4| = 2 \, |z - 1|$. Sketch the locus and state its centre and radius.

Rewrite as $|z - 4|/|z - 1| = 2$, so $k = 2 \neq 1$: an Apollonius circle. Square: $|z - 4|^2 = 4 |z - 1|^2$.

Squaring removes both moduli at once. The condition $k \neq 1$ guarantees a circle (not a line), so it is worth squaring.

PROBLEM 3 · UNIT CIRCLE AND ROTATION

Show that if $z$ lies on the unit circle $|z| = 1$, then so does $w = e^{i\pi/3} \cdot z$. Describe the geometric effect of multiplying by $e^{i\pi/3}$.

Compute $|w|$: $|w| = |e^{i\pi/3} \cdot z| = |e^{i\pi/3}| \cdot |z|$.

The modulus of a product is the product of the moduli — this is the workhorse identity for problems on $|z| = 1$.

Fill the gap: The locus $|z - z_0| = r$ in the Argand plane is a circle with centre and radius . Its Cartesian equation (with $z_0 = a + ib$) is $(x - a)^2 + (y - b)^2 = r^2$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Misreading the sign inside the modulus

$|z + 3 - 2i|$ is NOT centred at $(3, -2)$. Rewrite as $|z - (-3 + 2i)|$ first — then the centre is $(-3, 2)$. Always pull the expression into $|z - z_0|$ form before reading off the centre.

Trap 02

Assuming a ratio of distances is a line

$|z - a|/|z - b| = k$ is a line ONLY when $k = 1$. For any other positive $k$ it is an Apollonius circle. Square both sides and you will see the $x^2 + y^2$ terms do not cancel — meaning circle, not line.

Trap 03

Forgetting to square before squaring

$|z - z_0| = r$ gives $(x - a)^2 + (y - b)^2 = r^2$ — the RHS is $r^2$, not $r$. The modulus equals $r$, so the squared modulus equals $r^2$. Dropping the square is the most common silly error in MEX-N1.

Did you get this? True or false: the Cartesian equation of $|z + 1 - i| = 2$ is $(x + 1)^2 + (y - 1)^2 = 4$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Find the centre and radius of the circle $|z - 5 + 3i| = 7$ and write the Cartesian equation.

Where does the circle $|z - 2i| = 3$ cross the real and imaginary axes? Sketch.

Find the Cartesian equation of the locus $|z - 1| = 2|z + 2|$. Identify the shape, centre, and radius.

If $z$ lies on the unit circle $|z| = 1$, show that $|2z - 1| = |2 - \bar z|$ where $\bar z$ is the complex conjugate of $z$.

Sketch the locus $|z - 3| = |z - 3i|$. (Hint: this is the limiting case $k = 1$ — what kind of locus is it?)

Odd one out: Three of these describe the same circle in the Argand plane. Which one does NOT?

Revisit your thinking

Earlier you predicted the centre, radius, and real-axis crossings of $|z - (2 + i)| = 3$.

The Cartesian form is $(x - 2)^2 + (y - 1)^2 = 9$ — a circle centre $(2, 1)$, radius $3$. Setting $y = 0$ gives $(x - 2)^2 = 8$, so the circle crosses the real axis at $x = 2 \pm 2\sqrt{2}$. The pattern to internalise: modulus equals constant means circle; the centre is whatever $z$ is being measured from; the radius is the constant on the right. Almost every MEX-N1 circle problem follows this template.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Sketch the locus of $z$ in the Argand plane satisfying $|z - 1 - i| = 2$. State the centre and radius. (2 marks)

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ApplyBand 43 marks

Q2. Find the Cartesian equation of the locus $|z - 2| = 3|z + 2|$ and identify the centre and radius. (3 marks)

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AnalyseBand 53 marks

Q3. Suppose $z$ lies on the unit circle $|z| = 1$. Show that $w = (1 + z)/(1 - z)$ is purely imaginary (assume $z \neq 1$). (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. Rewrite $|z - 5 + 3i| = |z - (5 - 3i)| = 7$. Centre $(5, -3)$, radius $7$. Cartesian: $(x - 5)^2 + (y + 3)^2 = 49$.

2. Centre $(0, 2)$, radius $3$. Real-axis crossings ($y = 0$): $x^2 + 4 = 9 \Rightarrow x = \pm\sqrt{5}$. Imaginary-axis crossings ($x = 0$): $(y - 2)^2 = 9 \Rightarrow y = 5$ or $y = -1$.

3. Square: $(x - 1)^2 + y^2 = 4[(x + 2)^2 + y^2]$. Expand and simplify: $3x^2 + 3y^2 + 18x + 15 = 0$, so $x^2 + y^2 + 6x + 5 = 0$. Complete the square: $(x + 3)^2 + y^2 = 4$. Circle, centre $(-3, 0)$, radius $2$.

4. On $|z| = 1$, $\bar z = 1/z$. So $2 - \bar z = 2 - 1/z = (2z - 1)/z$. Take modulus: $|2 - \bar z| = |2z - 1|/|z| = |2z - 1|/1 = |2z - 1|$.

5. $k = 1$, so the locus is the perpendicular bisector of the segment from $3$ to $3i$. Midpoint $(\tfrac{3}{2}, \tfrac{3}{2})$; segment gradient $-1$; bisector gradient $1$ through the midpoint. Equation $y = x$.

Q1 (2 marks): Centre $(1, 1)$, radius $2$ [1]. Cartesian $(x - 1)^2 + (y - 1)^2 = 4$, sketch correct [1].

Q2 (3 marks): Square both sides [1]. Expand and collect: $x^2 + y^2 + 5x + 4 = 0$ [1]. Complete the square: $(x + \tfrac{5}{2})^2 + y^2 = \tfrac{9}{4}$, centre $(-\tfrac{5}{2}, 0)$, radius $\tfrac{3}{2}$ [1].

Q3 (3 marks): On $|z| = 1$, $\bar z = 1/z$ [1]. Compute $\bar w = (1 + \bar z)/(1 - \bar z) = (z + 1)/(z - 1) = -(1 + z)/(1 - z) = -w$ [1]. $\bar w = -w$ implies $w$ is purely imaginary [1].

Boss battle · The Locus Cartographer

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Five timed questions on circles, Apollonius loci, and the unit circle. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering quick locus questions. Lighter alternative to the boss.

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Module overview · Extension 2