Loci

A locus (plural: loci) is the set of all points that satisfy a given condition. In coordinate geometry, loci produce equations of curves. This lesson covers the most common loci at A-Level and how to derive their equations, as required by the Edexcel 9MA0 specification.

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What Is a Locus?

A locus is a path traced out by a point that moves according to a specific rule or condition.

Examples of loci:

• The set of points equidistant from a fixed point → a circle.
• The set of points equidistant from two fixed points → the perpendicular bisector of the line segment joining them.
• The set of points equidistant from a fixed point and a fixed line → a parabola.

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Locus 1: Equidistant from a Fixed Point (Circle)

The set of all points at distance r from a fixed point C(a, b) is a circle:

(x − a)² + (y − b)² = r²

Example: Find the locus of points that are 4 units from (2, −3).

(x − 2)² + (y + 3)² = 16.

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Locus 2: Equidistant from Two Points (Perpendicular Bisector)

The set of all points equidistant from A(x₁, y₁) and B(x₂, y₂) is the perpendicular bisector of AB.

Method: Set the distance from P(x, y) to A equal to the distance from P to B:

√((x − x₁)² + (y − y₁)²) = √((x − x₂)² + (y − y₂)²)

Square both sides and simplify.

Example: Find the locus of points equidistant from A(1, 3) and B(5, 1).

(x − 1)² + (y − 3)² = (x − 5)² + (y − 1)²

Expand: x² − 2x + 1 + y² − 6y + 9 = x² − 10x + 25 + y² − 2y + 1

Simplify: −2x − 6y + 10 = −10x − 2y + 26

8x − 4y = 16

2x − y = 4, or y = 2x − 4.

We can verify: the midpoint of AB is (3, 2), and 2(3) − 4 = 2 ✓. The gradient of AB is (1 − 3)/(5 − 1) = −1/2, and the perpendicular bisector has gradient 2 ✓.

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Locus 3: Point Dividing Distances in a Ratio

The set of points P such that PA/PB = k (a constant ratio, k ≠ 1) is a circle called the Apollonius circle.

Example: Find the locus of points where PA/PB = 2, with A = (0, 0) and B = (6, 0).

PA = 2PB, so PA² = 4PB²:

x² + y² = 4((x − 6)² + y²) x² + y² = 4x² − 48x + 144 + 4y² 0 = 3x² − 48x + 3y² + 144 x² − 16x + y² + 48 = 0 (x − 8)² + y² = 64 − 48 = 16

This is a circle with centre (8, 0) and radius 4.

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Locus 4: Condition Involving Gradients

Example: A point P(x, y) moves so that the line from A(0, 0) to P is perpendicular to the line from B(6, 0) to P. Find the locus of P.

Gradient of AP = y/x. Gradient of BP = y/(x − 6).

For perpendicularity: (y/x) × (y/(x − 6)) = −1

y² = −x(x − 6) = −x² + 6x

x² + y² − 6x = 0

(x − 3)² + y² = 9.

This is a circle with centre (3, 0) and radius 3 — the circle on diameter AB (Thales' theorem).

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Loci from Parametric Equations

Sometimes the locus is best described parametrically.

Example: A point P moves so that its x-coordinate is 3cos(θ) and its y-coordinate is 2sin(θ). What is the locus?

Eliminating θ: (x/3)² + (y/2)² = 1, which is an ellipse.

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Deriving the Locus Equation: General Strategy

1. Let P(x, y) be a general point on the locus.
2. Translate the geometric condition into an algebraic equation involving x and y.
3. Simplify the equation to a recognisable form (circle, line, parabola, etc.).
4. State the type of curve and any restrictions.

Exam Tip: Always square both sides when equating distances — this removes the square roots. Be careful to expand fully and simplify. The resulting equation should be in a standard recognisable form.

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Key Terms

Term	Definition
Locus	The set of all points satisfying a given geometric condition
Perpendicular bisector	The locus of points equidistant from two given points
Apollonius circle	The locus of points whose distances from two fixed points are in a constant ratio
Thales' theorem	An angle inscribed in a semicircle is a right angle

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Summary

• A locus is the path of a point satisfying a condition.
• Key loci include circles (equidistant from a point), perpendicular bisectors (equidistant from two points), and Apollonius circles (distance ratio).
• To find the equation: let P = (x, y), write the condition algebraically, and simplify.
• Always verify your answer by checking it satisfies the original condition at specific points.

A-Level Deep Dive: Loci

Spec mapping

Edexcel 9MA0-01 specification section 4 — Coordinate geometry in the (x, y) plane: loci appear as solution sets of geometric conditions, most commonly expressed in Cartesian form. The defining idea — a locus is the set of all points whose coordinates satisfy a stated geometric or algebraic condition — sits inside section 4 alongside straight lines, circles and parametric curves. Loci are not given a separate sub-strand on the 9MA0 specification page, but they are the conceptual glue tying section 4 together: a circle is the locus of points at fixed distance from a centre; the perpendicular bisector of $AB$AB is the locus of points equidistant from $A$A and $B$B; a parabola (treated only briefly in 9MA0 but explicitly in Further Maths 9FM0) is the locus equidistant from a focus and a directrix.

Synoptic mentions span the specification: section 2 (Algebra and functions) supplies the algebraic manipulation needed to convert "distance equals distance" into a Cartesian equation; section 3 (Sequences, series, inequalities) appears whenever a locus is presented as a region (an inequality) rather than a curve; section 4 (Coordinate geometry) treats circles as the canonical locus; and section 11 (Vectors) provides the alternative position-vector formulation $|\mathbf{r} - \mathbf{a}| = k$∣r−a∣=k for the same circle. Beyond 9MA0, Further Maths 9FM0 (Core Pure) treats Argand-diagram loci such as $|z - 1| = 2$∣z−1∣=2 (a circle in the complex plane), $\arg(z - a) = \theta$arg(z−a)=θ (a half-line), and $|z - a| = |z - b|$∣z−a∣=∣z−b∣ (a perpendicular bisector). Single-tier 9MA0 candidates do not need the complex-number version, but examiners expect them to recognise that the Cartesian and Argand formulations are the same locus dressed in different notation.

The Edexcel formula booklet gives the circle equation $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2 but does not give the perpendicular-bisector equation, the Apollonius-circle equation, or any locus identity. Candidates derive these every time from first principles using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$d=(x2​−x1​)2+(y2​−y1​)2​.

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Worked example with full mark scheme

Question (8 marks):

The point $P(x, y)$P(x,y) moves so that its distance from the point $A(1, 3)$A(1,3) is equal to its distance from the line $y = -1$y=−1.

(a) Show that the locus of $P$P has equation $(x - 1)^2 = 8(y - 1)$(x−1)2=8(y−1). (5)

(b) Hence state the coordinates of the vertex of this locus and identify the type of curve. (3)

Solution with mark scheme:

(a) Step 1 — write the two distances algebraically.

The distance from $P(x, y)$P(x,y) to $A(1, 3)$A(1,3) is:

$PA = \sqrt{(x - 1)^2 + (y - 3)^2}$PA=(x−1)2+(y−3)2​

The (perpendicular) distance from $P(x, y)$P(x,y) to the horizontal line $y = -1$y=−1 is:

$d = |y - (-1)| = |y + 1|$d=∣y−(−1)∣=∣y+1∣

M1 — both distances written correctly using the distance formula and the perpendicular-distance-to-a-horizontal-line interpretation. A common slip here is to write the distance to the line as $\sqrt{(x - x)^2 + (y + 1)^2}$(x−x)2+(y+1)2​, which collapses to $|y + 1|$∣y+1∣ but signals that the candidate has not yet realised the line is horizontal.

Step 2 — set the distances equal.

$\sqrt{(x - 1)^2 + (y - 3)^2} = |y + 1|$(x−1)2+(y−3)2​=∣y+1∣

M1 — equating the two distances. Note that for the equation to make sense as a distance, both sides must be non-negative; the modulus and the principal square root guarantee this.

Step 3 — square both sides to remove the radical.

$(x - 1)^2 + (y - 3)^2 = (y + 1)^2$(x−1)2+(y−3)2=(y+1)2

M1 — squaring both sides. Squaring is valid because both sides are non-negative; the modulus on the right squares away cleanly: $|y + 1|^2 = (y + 1)^2$∣y+1∣2=(y+1)2.

Step 4 — expand and simplify.

Expanding the $y$y-terms:

$(y - 3)^2 = y^2 - 6y + 9, \qquad (y + 1)^2 = y^2 + 2y + 1$(y−3)2=y2−6y+9,(y+1)2=y2+2y+1

Substituting and cancelling the common $y^2$y2:

$(x - 1)^2 + y^2 - 6y + 9 = y^2 + 2y + 1$(x−1)2+y2−6y+9=y2+2y+1

$(x - 1)^2 - 6y + 9 = 2y + 1$(x−1)2−6y+9=2y+1

$(x - 1)^2 = 8y - 8 = 8(y - 1)$(x−1)2=8y−8=8(y−1)

A1 — fully simplified locus equation $(x - 1)^2 = 8(y - 1)$(x−1)2=8(y−1).

A1 — presentation: equation written exactly in the printed form, not as an unsimplified equivalent like $(x - 1)^2 - 8y + 8 = 0$(x−1)2−8y+8=0.

(b) Step 1 — identify the curve.

The standard form $(x - h)^2 = 4p(y - k)$(x−h)2=4p(y−k) describes a parabola opening upwards with vertex $(h, k)$(h,k) and focus $(h, k + p)$(h,k+p).

B1 — recognising the equation $(x - 1)^2 = 8(y - 1)$(x−1)2=8(y−1) as a parabola.

Step 2 — read off the vertex.

Comparing $(x - 1)^2 = 8(y - 1)$(x−1)2=8(y−1) with $(x - h)^2 = 4p(y - k)$(x−h)2=4p(y−k) gives $h = 1$h=1, $k = 1$k=1, $4p = 8$4p=8 so $p = 2$p=2.

B1 — vertex at $(1, 1)$(1,1).

B1 — completing the verification: the focus is $(1, 3) = A$(1,3)=A and the directrix is $y = -1$y=−1, exactly the original condition. (This final mark rewards the candidate who closes the loop and verifies geometric consistency.)

Total: 8 marks (M3 A2 B3, split as shown).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): The points $A$A and $B$B have coordinates $(2, 0)$(2,0) and $(8, 0)$(8,0) respectively. The point $P$P moves in the $(x, y)$(x,y) plane such that $PA^2 + PB^2 = 50$PA2+PB2=50.

(a) Show that the locus of $P$P is a circle, and find its centre and radius. (4)

(b) Hence determine whether the point $Q(5, 4)$Q(5,4) lies inside, on, or outside this locus. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — writing $PA^2 = (x - 2)^2 + y^2$PA2=(x−2)2+y2 and $PB^2 = (x - 8)^2 + y^2$PB2=(x−8)2+y2 using the distance formula squared.
• M1 (AO1.1b) — substituting into the condition: $(x - 2)^2 + y^2 + (x - 8)^2 + y^2 = 50$(x−2)2+y2+(x−8)2+y2=50.
• A1 (AO1.1b) — expanding and simplifying to $2x^2 - 20x + 68 + 2y^2 = 50$2x2−20x+68+2y2=50, then $x^2 - 10x + 9 + y^2 = 0$x2−10x+9+y2=0, which completes-the-square to $(x - 5)^2 + y^2 = 16$(x−5)2+y2=16.
• A1 (AO2.5) — stating: centre $(5, 0)$(5,0), radius $4$4, identified explicitly as a circle.

(b)

• M1 (AO1.1b) — computing the distance from $Q(5, 4)$Q(5,4) to the centre $(5, 0)$(5,0): $d = \sqrt{0^2 + 4^2} = 4$d=02+42​=4.
• A1 (AO2.4) — concluding $d = r$d=r, so $Q$Q lies on the locus.

Total: 6 marks split AO1 = 4, AO2 = 2. Loci questions on Paper 1 sit roughly two-thirds AO1 (procedural — set up, expand, simplify) and one-third AO2 (interpretation — name the curve, locate centre/radius, classify position of test points). AO3 marks are rare on locus questions at single-tier 9MA0 because the geometric set-up is usually given.

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Synoptic links

Connects to:

1. Section 4 — Circles as loci: the equation $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2 is defined as the locus of points at distance $r$r from $(a, b)$(a,b). Every circle question is a locus question whose phrasing has been shortened. A* candidates write the unifying line "the circle is the locus of points satisfying $|\mathbf{r} - \mathbf{c}| = r$∣r−c∣=r" when asked to motivate the equation.
2. Section 4 — Perpendicular bisector as a locus: the perpendicular bisector of $AB$AB is the locus of points equidistant from $A$A and $B$B. Setting $PA = PB$PA=PB, squaring, and simplifying yields the linear equation $2(x_B - x_A)x + 2(y_B - y_A)y = |B|^2 - |A|^2$2(xB​−xA​)x+2(yB​−yA​)y=∣B∣2−∣A∣2 — which is exactly the perpendicular bisector. The "midpoint and gradient $-1/m$−1/m" recipe and the "set distances equal" derivation are the same line in disguise.
3. Further Maths 9FM0 — Argand-diagram loci: the complex-number condition $|z - 1| = 2$∣z−1∣=2 describes the locus of $z$z values at distance $2$2 from $1 + 0i$1+0i in the Argand plane — a circle. Substituting $z = x + iy$z=x+iy recovers $(x - 1)^2 + y^2 = 4$(x−1)2+y2=4. The condition $|z - a| = |z - b|$∣z−a∣=∣z−b∣ similarly recovers the perpendicular bisector. Single-tier 9MA0 candidates should recognise this as "the same locus with different clothing".
4. Conic sections (signposted to undergraduate analytic geometry): the parabola, ellipse and hyperbola are all loci defined by distance conditions — fixed distance from a point and a line (parabola), fixed sum of distances from two foci (ellipse), fixed difference of distances (hyperbola). 9MA0 treats only circles fully, but the parabola appears in section 8 (parametric equations).
5. Section 2 — Algebra and rearranging to Cartesian form: every locus problem reduces to algebra after the geometric set-up. Squaring, expanding brackets, completing the square, and identifying standard forms ($(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2 for a circle, $(x - h)^2 = 4p(y - k)$(x−h)2=4p(y−k) for a parabola) are pure section-2 manipulations. Locus questions thus test algebraic fluency through a geometric lens.

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Mark-scheme literacy

Locus questions on 9MA0 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–70%	Writing distances using the distance formula, squaring, expanding brackets, simplifying to a Cartesian form, completing the square
AO2 (reasoning / interpretation)	25–35%	Identifying the curve type (circle, parabola, line), reading off centre/radius/vertex, justifying the squaring step (both sides non-negative), classifying a test point as inside/on/outside
AO3 (problem-solving)	5–15%	Setting up the locus from a verbal condition with no algebraic scaffolding, combining a locus with another constraint (intersection problems)

Examiner-rewarded phrasing: "let $P = (x, y)$P=(x,y) be a general point on the locus"; "since both sides are non-negative, squaring is reversible"; "completing the square gives centre $(5, 0)$(5,0) and radius $4$4, so the locus is a circle"; "the locus is the set of all points satisfying ...". Phrases that lose marks: stating the curve type without justification ("it's a circle" with no derivation); writing the final equation without simplification (leaving $(x - 1)^2 - 8y + 8 = 0$(x−1)2−8y+8=0 when the printed answer is $(x - 1)^2 = 8(y - 1)$(x−1)2=8(y−1)); failing to reject extraneous solutions when the original condition imposes a sign constraint.

A specific Edexcel pattern to watch: questions that say "show that the locus is a circle and find its centre and radius" require both parts — a candidate who simplifies to $(x - 5)^2 + y^2 = 16$(x−5)2+y2=16 but stops there has shown the equation but not explicitly identified centre, radius, and type. The single A1 for "circle, centre $(5, 0)$(5,0), radius $4$4" is reserved for the candidate who states all three.

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Grade-band model answers

3-mark question

Question: The point $P(x, y)$P(x,y) is equidistant from $(0, 0)$(0,0) and $(6, 0)$(6,0). State the equation of the locus of $P$P and identify the geometric figure it represents.

Grade C response (~210 words):

If $P(x, y)$P(x,y) is equidistant from $O(0, 0)$O(0,0) and $A(6, 0)$A(6,0), then $PO = PA$PO=PA.

Squaring: $x^2 + y^2 = (x - 6)^2 + y^2$x2+y2=(x−6)2+y2.

Expanding: $x^2 + y^2 = x^2 - 12x + 36 + y^2$x2+y2=x2−12x+36+y2.

Cancelling $x^2$x2 and $y^2$y2: $0 = -12x + 36$0=−12x+36, so $12x = 36$12x=36, giving $x = 3$x=3.

So the locus is the line $x = 3$x=3. This is the perpendicular bisector of the segment from $O$O to $A$A.

Examiner commentary: Full marks (3/3). The candidate correctly equates the squared distances, simplifies cleanly, and identifies the locus geometrically as the perpendicular bisector. Working is brief but every step is justified. Squaring before equating saves time over taking square roots and squaring later — a small efficiency that A* candidates internalise. The phrase "perpendicular bisector" connects the algebraic answer to the geometric meaning, which is exactly the AO2 reasoning examiners reward. A weaker candidate would stop at $x = 3$x=3 without naming the curve, losing the interpretation mark on a longer question even though the algebra is correct.