OCR A Level Mathematics A: Straight lines and circles (H240)

This guide covers straight lines and circles in OCR A Level Mathematics A (H240), as set out in the OCR specification (Version 3.1, August 2026): content references 1.03a to 1.03f, within 1.03 Coordinate Geometry in the x-y plane. The specification lists all of this as Stage 1 content, the stage common to AS Level (H230) and A Level, so none of it is A Level only. Parametric equations (1.03g and 1.03h) are not covered here.

Use it with the revision notes and the practice set. For the wider course see the Pure Mathematics guide, the course hub and the checklist. The free diagnostics show where to start.

Scope at a glance

Ref	Skills	Stage
1.03a	Use y = mx + c, y - y1 = m(x - x1) and ax + by + c = 0; draw a line and form its equation from a graph, a gradient and a point, or two points; find a midpoint, a distance and an intersection	1
1.03b	Use the gradient conditions for parallel and perpendicular lines	1
1.03c	Use straight line models in context, including average rates of change	1
1.03d	Use the equation (x - a)² + (y - b)² = r²; draw a circle from its equation or form it from centre and radius	1
1.03e	Complete the square to find a circle’s centre and radius	1
1.03f	Use three circle properties in coordinate problems; decide whether a line and a circle, or two circles, intersect	1

1.03a Equations of straight lines

Three forms are named in the specification.

Form	Use it when
y = mx + c	you know the gradient m and the y-intercept c
y - y1 = m(x - x1)	you know the gradient and one point (x1, y1)
ax + by + c = 0	an answer is wanted with integer coefficients

The gradient between (x1, y1) and (x2, y2) is m = (y2 - y1)/(x2 - x1). Two further results come up constantly:

• Midpoint: ((x1 + x2)/2, (y1 + y2)/2)
• Distance: √[(x2 - x1)² + (y2 - y1)²]

To draw a line, find two points, usually the intercepts: put x = 0, then y = 0. To form an equation from a graph, read off two clear points (or the gradient and one point) and continue as for two points.

Worked example 1. Find the equation of the line through A(-2, 7) and B(4, -5), the midpoint of AB and the length AB.

m = (-5 - 7)/(4 - (-2)) = -12/6 = -2
y - 7 = -2(x + 2)  →  y = -2x + 3  →  2x + y - 3 = 0
midpoint = ((-2 + 4)/2, (7 - 5)/2) = (1, 1)
AB = √(6² + 12²) = √180 = 6√5

Check: the midpoint (1, 1) gives 2 + 1 - 3 = 0, so it lies on the line.

Worked example 2. A line has gradient 3/5 and passes through (10, -1). Write it as ax + by + c = 0 with integer coefficients, then find where it meets x + 2y + 3 = 0.

y + 1 = (3/5)(x - 10)
5y + 5 = 3x - 30
3x - 5y - 35 = 0

Intersection: from the second line x = -3 - 2y. Substitute:

3(-3 - 2y) - 5y = 35
-9 - 11y = 35  →  y = -4,  x = -3 + 8 = 5

The lines meet at (5, -4). Always check by substituting into both equations.

1.03b Parallel and perpendicular lines

• Parallel lines: m1 = m2
• Perpendicular lines: m1 × m2 = -1

So a line with gradient 3/5 has perpendicular gradient -5/3: flip the fraction and change the sign. To read a gradient from ax + by + c = 0, rearrange to y = mx + c first. It is -a/b.

Worked example 3. Find the perpendicular bisector of P(-6, 1) and Q(4, 7).

gradient PQ = (7 - 1)/(4 + 6) = 6/10 = 3/5
midpoint = (-1, 4)
perpendicular gradient = -5/3
y - 4 = -(5/3)(x + 1)
3y - 12 = -5x - 5
5x + 3y - 7 = 0

Perpendicular gradients also prove right angles. If the gradients of two sides of a triangle multiply to -1, the angle between them is 90°.

1.03c Straight line models

A linear model assumes a constant rate of change. The gradient is that rate, with units “units of y per unit of x”. The intercept is the starting value. The specification says these problems may be set in realistic contexts, including average rates of change.

Worked example 4. A tank drains steadily. After 4 minutes it holds 310 litres, and after 10 minutes it holds 238 litres. Model the volume V litres after t minutes.

gradient = (238 - 310)/(10 - 4) = -12
V - 310 = -12(t - 4)  →  V = 358 - 12t

• The gradient -12 means the tank loses 12 litres each minute.
• The intercept 358 is the volume at t = 0.
• The tank is empty when 358 - 12t = 0, so t = 179/6 ≈ 29.8 minutes.

A limitation to state: the model is only valid while V ≥ 0, and it assumes the flow never changes. Real drains often slow as the level falls.

1.03d The equation of a circle

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

(x - a)² + (y - b)² = r²

Note the signs: (x + 3)² means a = -3. A point is inside the circle if (x - a)² + (y - b)² < r², on it if equal, and outside if greater.

Worked example 5. A circle has centre (3, -2) and passes through (5, 3). Find its equation and where it crosses the axes.

r² = (5 - 3)² + (3 + 2)² = 4 + 25 = 29
(x - 3)² + (y + 2)² = 29

Where x = 0: 9 + (y + 2)² = 29, so (y + 2)² = 20 and y = -2 ± 2√5. Where y = 0: (x - 3)² + 4 = 29, so x - 3 = ±5 and x = -2 or x = 8.

To draw it, plot the centre, then mark points a distance r above, below, left and right of it, and join them in a smooth curve. Here r = √29 ≈ 5.4.

1.03e Completing the square for circles

When the equation is expanded, group x terms and y terms and complete the square on each.

Worked example 6. Find the centre and radius of x² + y² - 8x + 10y + 5 = 0.

(x² - 8x) + (y² + 10y) = -5
(x - 4)² - 16 + (y + 5)² - 25 = -5
(x - 4)² + (y + 5)² = 36

Centre (4, -5), radius 6.

Worked example 7. Find the centre and radius of 2x² + 2y² - 12x + 4y - 9 = 0.

Divide by 2 first so the coefficients of x² and y² are 1:

x² + y² - 6x + 2y - 9/2 = 0
(x - 3)² - 9 + (y + 1)² - 1 = 9/2
(x - 3)² + (y + 1)² = 29/2

Centre (3, -1), radius √(29/2).

When is it a circle? In x² + y² - 4x + 6y + k = 0 the completed form is (x - 2)² + (y + 3)² = 13 - k. The right-hand side is r², so it must be positive: k < 13. If it equals 0 the “circle” is a single point; if negative, nothing is plotted.

1.03f Circle properties and intersections

The specification names three properties for use in coordinate geometry problems:

1. Any angle drawn in a semicircle is a right angle.
2. A line from the centre at right angles to a chord cuts that chord in half.
3. The radius at a point on the circumference is perpendicular to the tangent at that point.

The normal to a circle at a point is the line through that point and the centre.

Worked example 8 (semicircle). A(0, 1) and B(6, 5) are ends of a diameter. Show that P(5, 0) is on the circle.

gradient PA = (0 - 1)/(5 - 0) = -1/5
gradient PB = (0 - 5)/(5 - 6) = 5
product = -1

So angle APB is 90°, and P lies on the circle with diameter AB. As a check, the centre is (3, 3) and r² = 13, and (5 - 3)² + (0 - 3)² = 13.

Worked example 9 (chord). The line y = x + 1 meets (x - 1)² + (y + 2)² = 40 at two points. Find the midpoint and length of the chord.

The centre is (1, -2). The perpendicular from the centre has gradient -1: y + 2 = -(x - 1), so y = -x - 1. It meets the chord where x + 1 = -x - 1, giving the midpoint M(-1, 0).

CM² = (1 + 1)² + (-2 - 0)² = 8
half chord = √(40 - 8) = √32
chord = 2√32 = 8√2

Substitution confirms it: x² + 2x - 15 = 0 gives x = 3 or -5, so the ends are (3, 4) and (-5, -4), with midpoint (-1, 0).

Worked example 10 (tangent and normal). Find the tangent and normal to (x + 2)² + (y - 1)² = 50 at (5, 2).

First check the point: 49 + 1 = 50.

radius gradient = (2 - 1)/(5 + 2) = 1/7
tangent gradient = -7
tangent: y - 2 = -7(x - 5)  →  7x + y - 37 = 0
normal:  y - 2 = (1/7)(x - 5)  →  x - 7y + 9 = 0

Does a line meet a circle? Substitute the line into the circle to get a quadratic in one variable, then look at the discriminant b² - 4ac: positive means two points, zero means tangent, negative means no intersection.

Worked example 11. For which values of k does y = x + k meet x² + y² = 18?

x² + (x + k)² = 18
2x² + 2kx + k² - 18 = 0
discriminant = 4k² - 8(k² - 18) = 144 - 4k²

Two points when 144 - 4k² > 0, so -6 < k < 6. Tangent when k = ±6. No intersection when |k| > 6.

Two circles. Compare the distance d between centres with the radii r1 and r2. They meet at two points if |r1 - r2| < d < r1 + r2. They touch externally if d = r1 + r2, touch internally if d = |r1 - r2|, and do not meet if d is larger than r1 + r2 or smaller than |r1 - r2|.

Example: x² + y² = 16 and (x - 6)² + (y - 2)² = 9 have d = √40 ≈ 6.32, r1 + r2 = 7 and |r1 - r2| = 1. Since 1 < 6.32 < 7, they meet at two points.

Common errors

• Writing the perpendicular gradient as the negative only, or the reciprocal only. You need both changes.
• Reading the centre of (x + 3)² + (y - 4)² = 25 as (3, -4). It is (-3, 4).
• Giving the radius as r² or forgetting the square root: (x - 1)² + (y - 2)² = 36 has radius 6, not 36.
• Not dividing by the coefficient of x² and y² before completing the square.
• Finding the tangent gradient from the radius but using the centre as the point, rather than the point of contact.
• Leaving a gradient model without units, or applying it far outside the data.

Where next

Try the practice questions and keep the revision notes for the final weeks.

Official syllabus

Cambridge OCR Level 3 Advanced GCE in Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by Cambridge OCR: section 1.03, Coordinate Geometry in the x-y plane, references 1.03a to 1.03f.