AQA A-Level Mathematics: C: Coordinate geometry in the (x, y) plane (7357)

This study guide teaches Section C: Coordinate geometry in the (x, y) plane of the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams from June 2018 onwards. It covers every content reference in the section, C1 to C4: straight lines, circles, and parametric equations. Section C is listed under Paper 1, and Papers 2 and 3 can assess any Paper 1 content as well, so these skills can earn marks on all three papers. A calculator is required in every 7357 assessment, but most coordinate geometry marks come from exact algebra, so show it.

Use it with the revision notes and the practice questions. The course hub is at /boards/aqa/a-level/mathematics/ and the printable checklist at /checklists/aqa/a-level/mathematics/.

What this section covers

Ref	What you must be able to do
C1	Use the equation of a straight line, including y − y₁ = m(x − x₁) and ax + by + c = 0; use the gradient conditions for parallel and perpendicular lines; use straight-line models in context
C2	Use the circle (x − a)² + (y − b)² = r²; complete the square to find the centre and radius; use three circle properties (angle in a semicircle, perpendicular from the centre to a chord, radius and tangent)
C3	Use parametric equations of curves and convert between Cartesian and parametric forms
C4	Use parametric equations in modelling in a variety of contexts

Two results here are in Appendix B of the specification, which lists formulae you must recall because they are not given in the exam: the line y − y₁ = m(x − x₁), and the condition m₁m₂ = −1 for perpendicular lines.

C1: Straight lines

The gradient of the line through (x₁, y₁) and (x₂, y₂) is

m = (y₂ − y₁)/(x₂ − x₁)

You will meet three forms of the equation:

• y = mx + c — gradient m, y-intercept c. Best for reading off the gradient.
• y − y₁ = m(x − x₁) — gradient m through (x₁, y₁). Fastest way to write an equation.
• ax + by + c = 0 — a general form. Questions often ask for a, b and c as integers.

To find the gradient from ax + by + c = 0, rearrange to y = −(a/b)x − c/b. The gradient is −a/b.

Parallel and perpendicular lines. Parallel lines have equal gradients: m₁ = m₂. Perpendicular lines have m₁m₂ = −1, so the perpendicular gradient is the negative reciprocal: −1/m. A gradient of −2/3 becomes 3/2.

Worked example 1

A is (−2, 5) and B is (4, 1). Find the equation of line AB in the form ax + by + c = 0, where a, b and c are integers, then find the perpendicular bisector of AB.

Gradient AB = (1 − 5)/(4 − (−2)) = −4/6 = −2/3
y − 5 = −(2/3)(x + 2)
3y − 15 = −2x − 4
2x + 3y − 11 = 0

The perpendicular bisector passes through the midpoint of AB at right angles to it.

Midpoint = ((−2 + 4)/2, (5 + 1)/2) = (1, 3)
Perpendicular gradient = 3/2
y − 3 = (3/2)(x − 1)
2y − 6 = 3x − 3
3x − 2y + 3 = 0

Check: (1, 3) gives 3 − 6 + 3 = 0. The length AB = √(6² + 4²) = √52 = 2√13, which you will need when a circle has AB as a diameter.

Straight-line models

C1 also asks you to use straight lines as models. The gradient is a rate of change and the intercept is a starting value. You must interpret both in context and say where the model stops working.

Worked example 2

A candle is 24 cm tall when it is lit. After 3 hours it is 18.6 cm tall. Assume the height h cm falls linearly with time t hours.

Gradient = (18.6 − 24)/3 = −1.8
h = 24 − 1.8t

The gradient −1.8 means the candle gets 1.8 cm shorter each hour. The intercept 24 is the height when lit. The candle has burnt down when h = 0, so t = 24/1.8 = 40/3 hours (13 hours 20 minutes). The model is only valid for 0 ≤ t ≤ 40/3; it would give negative heights after that. A refinement might allow for the wick and the base, which do not burn at the same rate.

C2: Circles

A circle with centre (a, b) and radius r has equation

(x − a)² + (y − b)² = r²

This is Pythagoras’ theorem: every point (x, y) on the circle is a distance r from (a, b). Watch the signs. (x + 4)² means a = −4.

Completing the square

If the equation is expanded, complete the square in x and in y separately.

Worked example 3

Find the centre and radius of x² + y² − 6x + 10y + 9 = 0.

(x² − 6x) + (y² + 10y) + 9 = 0
(x − 3)² − 9 + (y + 5)² − 25 + 9 = 0
(x − 3)² + (y + 5)² = 25

Centre (3, −5), radius 5. If the right-hand side comes out zero or negative, the equation does not describe a circle.

Property 1: the angle in a semicircle is a right angle

If PQ is a diameter and R is any other point on the circle, angle PRQ = 90°. You can use it both ways: a right angle at R tells you PQ is a diameter, so the centre is the midpoint of PQ.

Worked example 4

P(1, 7) and Q(7, −1) are the ends of a diameter. Find the circle’s equation and show that angle PRQ is a right angle, where R is (8, 6).

Centre = midpoint of PQ = (4, 3)
PQ = √(6² + 8²) = √100 = 10, so r = 5
Circle: (x − 4)² + (y − 3)² = 25
R: (8 − 4)² + (6 − 3)² = 16 + 9 = 25, so R is on the circle
Gradient PR = (6 − 7)/(8 − 1) = −1/7
Gradient QR = (6 + 1)/(8 − 7) = 7
Product = −1, so PR is perpendicular to QR

Property 2: the perpendicular from the centre to a chord bisects the chord

So the centre always lies on the perpendicular bisector of any chord. This lets you find a centre from two points on the circle.

Worked example 5

A circle passes through A(2, 1) and B(6, 5). Its centre lies on y = x − 3. Find its equation.

Midpoint AB = (4, 3); gradient AB = 4/4 = 1
Perpendicular bisector: y − 3 = −1(x − 4), so y = −x + 7
Centre: −x + 7 = x − 3, so x = 5, y = 2
r² = (2 − 5)² + (1 − 2)² = 9 + 1 = 10
Circle: (x − 5)² + (y − 2)² = 10

Check with B: (6 − 5)² + (5 − 2)² = 1 + 9 = 10.

The same property gives chord lengths. If a chord is a distance d from the centre, half the chord is √(r² − d²).

Property 3: the radius is perpendicular to the tangent

At a point on the circle, the tangent is at right angles to the radius. Find the gradient of the radius, then take its negative reciprocal.

Worked example 6

Find the tangent to (x − 3)² + (y + 5)² = 25 at (6, −1).

Check: (6 − 3)² + (−1 + 5)² = 9 + 16 = 25
Gradient of radius = (−1 − (−5))/(6 − 3) = 4/3
Gradient of tangent = −3/4
y + 1 = −(3/4)(x − 6)
4y + 4 = −3x + 18
3x + 4y − 14 = 0

Lines meeting circles

To find where a line meets a circle, substitute the line into the circle and solve the quadratic. Its discriminant tells you how many points there are: two (b² − 4ac > 0), one, so the line is a tangent (= 0), or none (< 0). The quadratics study guide covers the discriminant itself.

Worked example 7

Find where y = x + 4 meets (x − 2)² + (y − 1)² = 25, and the length of the chord.

(x − 2)² + (x + 3)² = 25
2x² + 2x + 13 = 25
x² + x − 6 = 0
(x + 3)(x − 2) = 0, so x = −3 or x = 2
Points (−3, 1) and (2, 6)
Chord = √(5² + 5²) = 5√2

C3: Parametric equations

A parametric curve gives x and y separately in terms of a third variable, the parameter (often t or θ). Each value of the parameter gives one point.

Parametric to Cartesian: eliminate the parameter. Either make t the subject of one equation and substitute, or use an identity such as sin²θ + cos²θ ≡ 1 or cos 2θ ≡ 2cos²θ − 1.

Cartesian to parametric: choose a substitution that fits. The circle (x − a)² + (y − b)² = r² has parametric form x = a + r cos θ, y = b + r sin θ, for 0 ≤ θ < 2π.

Worked example 8

Find the Cartesian equation of x = 1 + 2t, y = t² − 3.

t = (x − 1)/2
y = ((x − 1)/2)² − 3
y = (x² − 2x − 11)/4

Worked example 9

A curve has x = 2cos θ, y = cos 2θ, 0 ≤ θ ≤ π. Find its Cartesian equation and state the possible values of x.

cos 2θ = 2cos²θ − 1 and cos θ = x/2
y = 2(x/2)² − 1 = x²/2 − 1

Since −1 ≤ cos θ ≤ 1, the curve is only the part of y = x²/2 − 1 with −2 ≤ x ≤ 2. Always state this restriction: the Cartesian equation alone describes a longer curve.

Parametric curves are differentiated using dy/dx = (dy/dt) ÷ (dx/dt); that method is content reference G5, covered in the differentiation study guide.

C4: Parametric equations in modelling

In a model, the parameter is usually time. The parametric form tells you where something is when, which a Cartesian equation cannot.

Worked example 10

A ball is thrown from a point 1.5 m above level ground. Its position t seconds later is modelled by x = 12t, y = 1.5 + 14t − 4.9t², where x and y are in metres, x horizontal and y vertical.

(a) Find the greatest height. The y-equation is a quadratic in t; complete the square or use its symmetry. The maximum is at t = 14/9.8 = 10/7 s.

y = 1.5 + 14(10/7) − 4.9(10/7)² = 1.5 + 20 − 10 = 11.5 m

(b) Find how far the ball travels horizontally before it lands. Solve y = 0 and take the positive root.

4.9t² − 14t − 1.5 = 0
t = (14 + √(196 + 29.4))/9.8 = 2.96 s (3 s.f.)
x = 12 × 2.960... = 35.5 m (3 s.f.)

Use the unrounded t to find x. The negative root is rejected because the model starts at t = 0.

(c) Limitations: the model ignores air resistance and treats the ball as a particle. It only holds from release until the ball lands, 0 ≤ t ≤ 2.96. The overarching themes guide explains the modelling cycle (OT3) that this question uses.

Common errors

• Writing the gradient of ax + by + c = 0 as a/b or −b/a. It is −a/b.
• Taking the perpendicular gradient as 1/m or −m instead of −1/m.
• Reading the centre of (x + 4)² + (y − 1)² = 25 as (4, −1). It is (−4, 1).
• Giving r² as the radius: (x − 1)² + y² = 49 has radius 7, not 49.
• Forgetting to subtract both completed-square constants before reading off r².
• Using the gradient of the radius as the tangent gradient.
• Not giving integer a, b, c when the question asks for ax + by + c = 0 “where a, b and c are integers”.
• Leaving out the x-range after eliminating a trigonometric parameter.
• Rounding t before using it to find x in a parametric model.

Next steps

Condense this into the revision notes, then test yourself on the practice questions. For exam technique across all three papers, read the 7357 exam preparation guide. You can also try one of the free diagnostics.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3, 31 January 2018 (A-level exams June 2018 onwards), section 3.4 C: Coordinate geometry in the (x, y) plane, content references C1 to C4. Published by AQA.