AQA A-Level Mathematics Paper 2, June 2025: Question 9

9 marks · Medium difficulty · Multi-step Problem

Analyze a circle and a modulus function intersecting at points A, B, and D, determining coordinates, finding an unknown coordinate, calculating angle ADB in radians, and finding the minor arc length AB.

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Question

Question text

9 A circle with centre C has equation

(x – 12)2 + (y – 2)2 = 100

The graph with equation

y = │3x – 36│ – 8

intersects the circle at the points A, B and D as shown in the diagram.

A B

O x

Point D is vertically below point C

9 (a) State the coordinates of C

[1 mark]

9 (b) State the coordinates of D

[1 mark]

9 (c) The coordinates of A are (a,10)

Find the value of a

(14)Fully justify your answer.

[3 marks]

9 (d) (i) Find, in radians, the angle ADB

Give your answer to three significant figures.

[2 marks]

9 (d) (ii) Hence or otherwise find the length of the minor arc AB

Give your answer to three significant figures.

[2 marks]

Mark scheme

Show the mark scheme

Q Marking instructions AO Marks Typical solution

9(a) States (12, 2) 1.1b B1 (12, 2)

Subtotal 1

9(b) States (12, –8) 2.2a R1 (12, –8)

Subtotal 1

Forms an equation for x or a by

9(c) Substituting y = 10 into 3.1a M1

( )2 ( )2

x −12 + y − 2 = 100 or

y = 3x −36 −8

Substitutes (3x −36)−8for y

in the circle equation

Forms a right-angled triangle

with a hypotenuse of 10 and a

shorter side of 8

and 2 2

uses an appropriate process to (a −12) + (10 − 2) =100

find the length of the third side a = 6 or 18

PI by a 6, 8, 10 right-angled

triangle 6 18

Solves their equation to find a 1.1a M1 a = 6

value of x or a

Identifies 6, 8 and 10 as a

Pythagorean triple

PI by a 6, 8, 10 right-angled

triangle

Completes reasoned argument 2.1 R1

to obtain a = 6 with a fully

correct solution

If a 6, 8, 10 approach is used we

must see 12 – 6 = a, 12 – 6 = 6

or 12 – a = 6

Accept (6, 10) – A-LEVEL MATHEMATICS – 7357/2 –

NMS scores M0M0R0

Subtotal 3

Uses an appropriate

9(d)(i) trigonometric equation to find 3.1a M1 6

1 tan =

angle ADB or ADB such as 2 18

2 = 0 644.

arctan or arctan OE

18 8

PI by AWRT 18 4.o or 36 9.o

Obtains AWRT 0.644 1.1b A1

12 Subtotal 2

Uses l = r with either

9(d)(ii) r = 10 or = 2 their 0 644. 3.1a M1 l = 10 (2 0 644.)

OE = 12 9.

Condone angle in degrees

Obtains AWFW [12.8,12.9] 1.1b A1

Subtotal 2

Question 9 Total 9

How to answer it

Circle Geometry, Modulus Functions & Arc Length

📌 What this question tests

This question brings together core Pure Mathematics concepts: extracting the centre and radius from a standard circle equation (x - h)² + (y - k)² = r² , finding coordinates using symmetry and geometry, solving intersection problems involving modulus/quadratic equations with full geometric justification, and applying trigonometry and circle theorems to determine angles and arc lengths using radians ( s = rθ ).

Part (a) — [1 Mark]

Coordinates of Circle Centre C

Identifying centre from standard equation of a circle

✅ Correct Answer

The coordinates are (12, 2).

💡 Key Knowledge

The standard equation of a circle with centre (h, k) and radius r is:

(x - h)² + (y - k)² = r²

Comparing with (x - 12)² + (y - 2)² = 100 gives h = 12 , k = 2 , and r = √100 = 10 .

Mark Scheme: B1 for stating (12, 2) .

Part (b) — [1 Mark]

Coordinates of Intersection Point D

Using vertical alignment and circle radius or modulus vertex

✅ Correct Answer

The coordinates are (12, -8).

📐 Calculations

Method 1 (Geometric):

Point D lies vertically below C , so its x-coordinate is identical: x = 12 .
D lies on the circle of radius r = 10 directly below centre (12, 2) :
y = 2 - 10 = -8 .

Method 2 (Modulus Vertex):

Vertex of y = |3x - 36| - 8 occurs when 3x - 36 = 0 ⇒ x = 12 .
When x = 12 , y = 0 - 8 = -8 .

Mark Scheme: R1 for deducing and stating (12, -8) .

Part (c) — [3 Marks]

Finding the Value of a (Coordinates of A)

Solving an equation and providing a reasoned justification

📐 Step-by-Step Calculation

Step 1: Point A(a, 10) lies on the circle. Substitute x = a and y = 10 into the circle equation:

(a - 12)² + (10 - 2)² = 100

(a - 12)² + 8² = 100

(a - 12)² + 64 = 100

(a - 12)² = 36

Step 2: Solve for a :

a - 12 = ±6 ⇒ a = 12 - 6 = 6 or a = 12 + 6 = 18

Step 3: Reject the incorrect solution with full justification:

From the diagram, point A is the point to the left of the line of symmetry x = 12 (or a < 12 , or 6 < 18 , where 18 corresponds to B ).

Therefore, a = 6.

❌ Common Errors & Traps

Missing the justification: The question asks to "Fully justify your answer". Stating a = 6 without explaining why 18 was rejected loses the reasoning mark (R1).
No Method Shown (NMS): Writing down a = 6 by inspection or guessing scores 0/3.
Modulus branch ambiguity: If substituting into y = |3x - 36| - 8 , remember A is on the decreasing branch where 3x - 36 < 0 , so -(3a - 36) - 8 = 10 ⇒ -3a + 28 = 10 ⇒ a = 6 .

Mark Scheme:
M1 (3.1a): Sets up equation by substituting y = 10 into the circle or modulus equation, OR sets up a 6-8-10 right-angled triangle.
M1 (1.1a): Solves equation to obtain a = 6 or 18 .
R1 (2.1): Completes a reasoned argument (e.g. noting a < 12 , 6 < 18 , or 12 - 6 = 6 ) to conclude definitively that a = 6 .

Part (d)(i) — [2 Marks]

Finding Angle ADB in Radians

Right-angled trigonometry and symmetry

📐 Step-by-Step Calculation

Step 1: Set up a right-angled triangle using symmetry.

The line x = 12 is an axis of vertical symmetry through vertex D(12, -8) .

Let M(12, 10) be the midpoint of chord AB .

Horizontal distance from M to A : 12 - 6 = 6
Vertical distance from D to M : 10 - (-8) = 18

In right-angled triangle AMD , let angle ADM = θ / 2 :

tan(θ / 2) = 6 / 18 = 1 / 3

θ / 2 = arctan(1 / 3) ≈ 0.32175 rad

Step 2: Double the half-angle to get angle ADB:

θ = 2 × 0.32175... = 0.64350... rad

Rounding to 3 significant figures gives: 0.644 radians.

🧠 Alternative Geometry Method

Consider the angle at the centre C(12, 2) :

Horizontal distance from C to A : 6
Vertical distance from C to chord: 10 - 2 = 8
Half-angle at centre: tan(∠ACM) = 6 / 8
By the circle theorem: Angle at the centre is twice the angle at the circumference ( ∠ACB = 2 × ∠ADB ).
Hence, ∠ADB = ∠ACM = arctan(6 / 8) ≈ 0.6435 rad !

Mark Scheme:
M1 (3.1a): Uses an appropriate trigonometric equation to find angle ADB or ½ ADB , e.g. arctan(6/18) or arctan(6/8) (seen or implied by 18.4° or 36.9°).
A1 (1.1b): Obtains 0.644 (AWRT).

Part (d)(ii) — [2 Marks]

Length of Minor Arc AB

Applying circle theorems and arc length formula s = rθ

📐 Step-by-Step Calculation

Step 1: Find the subtended angle at the centre C.

Angle ADB is an inscribed angle subtending arc AB .

By circle theorems, the angle subtended at the centre C is twice the angle subtended at the circumference:

θ_centre = 2 × ∠ADB = 2 × 0.64350... = 1.2870... rad

Step 2: Apply the radian arc length formula.

From the circle equation r² = 100 ⇒ r = 10 :

l = r × θ_centre

l = 10 × 1.2870... = 12.870...

Rounding to 3 significant figures: 12.9 (or 12.88).

❌ Common Errors

Using angle ADB directly in s = rθ: Calculating 10 × 0.644 = 6.44 forgets that s = rθ requires the angle subtended at the centre of the circle, not at the circumference!
Degree-radian confusion: Leaving your calculator in degrees or mixing degree formulas with radians without converting.
Premature rounding: Rounding 0.644 too early: 10 × 2 × 0.644 = 12.88 → 12.9 . Always keep unrounded values stored in memory!

Mark Scheme:
M1 (3.1a): Uses l = rθ with radius r = 10 or angle at centre θ = 2 × (their 0.644) .
A1 (1.1b): Obtains 12.9 (accept any value in the range [12.8, 12.9]).

Topics

Pure Mathematics · B: Algebra and functions · C: Coordinate geometry in the (x, y) plane · E: Trigonometry

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.