In Year 10 you met the foundations of coordinate geometry — points, distances and the equation of a line. In Year 11, on the Higher tier, the topic returns at a much higher level: the circle enters the picture, and with it tangents, points of intersection, and combined questions that mix a line and a circle in the same diagram. The GCSE Higher question is almost never 'find the equation of the circle' on its own — it is always a chain of steps: find the centre and radius, draw a line in a given position, check where they meet, and prove a tangent. This guide takes you through every tool in the right order, with an emphasis on the connections between the line and the circle — exactly where the marks are lost in the exam.
Coordinate geometry is the chapter where algebra and geometry meet: every point is a pair of numbers, every line and every circle is an equation, and every geometric question — intersection, perpendicularity, distance, tangency — becomes an algebraic calculation. On GCSE Higher (specification reference A16, examined on any of the three papers) the equation of a circle is one of the questions you can genuinely 'close' — it is technical, predictable, and works to fixed patterns. The difference between a pupil who gets full marks on it and a pupil who gets half is not talent, but working order: knowing exactly which tool to pull out at each stage. In Year 10 you learnt points, distances and the equation of a line; here we assume you have all that, refresh it quickly, and then move to the core of the Year 11 Higher material — the circle, the tangent, and the combined problems that join a line and a circle in one diagram.
Points and line segments — distance and midpoint
The two most basic tools are the distance between two points and the midpoint of a line segment. Both come straight from Pythagoras' theorem, and both appear in almost every exam question on the topic.
Distance between two points A(x₁, y₁) and B(x₂, y₂): d = √[(x₂ − x₁)² + (y₂ − y₁)²]. Notice that the differences in the coordinates are squared, so it does not matter whether you subtract A from B or the other way round. Example: the distance between (1, 2) and (4, 6) is √[(4 − 1)² + (6 − 2)²] = √[9 + 16] = √25 = 5.
Midpoint of a segment — the mean of the coordinates. The midpoint of the segment with ends A(x₁, y₁) and B(x₂, y₂) is M((x₁ + x₂)/2, (y₁ + y₂)/2). Example: the midpoint of the segment from (2, 3) to (8, 1) is ((2 + 8)/2, (3 + 1)/2) = (5, 2). An important use: the centre of a circle is the midpoint of a diameter, so if you are given the two ends of a diameter — their midpoint is the centre.
The equation of a line and the gradient
The gradient m measures the steepness of the line — how much it rises vertically for every unit horizontally. Between two points: m = (y₂ − y₁)/(x₂ − x₁). A positive gradient = a rising line, a negative gradient = a falling line, a zero gradient = a horizontal line, and an undefined gradient (zero denominator) = a vertical line.
The most useful form of the equation in the exam is the point–gradient form: y − y₁ = m(x − x₁). If you are given a gradient m and a point (x₁, y₁) the line passes through — substitute and finish. Example: a line with gradient 2 through the point (3, 1): y − 1 = 2(x − 3), that is y = 2x − 5.
Finding the equation of a line through two points in two steps: (1) work out the gradient from the formula, (2) substitute the gradient and one of the points into the point–gradient form. Example: a line through (1, 2) and (4, 8). Gradient: m = (8 − 2)/(4 − 1) = 6/3 = 2. Equation: y − 2 = 2(x − 1), that is y = 2x.
Parallel and perpendicular lines
These two conditions are the beating heart of most combined problems, and especially of the tangent questions we will reach later.
Parallel lines — equal gradients: m₁ = m₂. Two lines are parallel if and only if they have the same gradient (and are not the same line). Example: y = 3x + 1 and y = 3x − 7 are parallel, because both have gradient 3.
Perpendicular lines — the product of the gradients is minus one: m₁ × m₂ = −1. In other words one gradient is the negative reciprocal of the other: m₂ = −1/m₁. Example: if a line has gradient 2, then any line perpendicular to it has gradient −1/2, because 2 × (−1/2) = −1. This is the most critical rule in tangent questions: the tangent to a circle is perpendicular to the radius at the point of contact.
| Relationship between the lines | Condition on the gradients | Example |
|---|---|---|
| Parallel | m₁ = m₂ | y = 4x and y = 4x + 9 |
| Perpendicular | m₁ × m₂ = −1 | gradient 3 and gradient −1/3 |
| Intersecting (general) | m₁ ≠ m₂ | y = x and y = 2x |
| The same line | same m and same intercept | y = 2x + 1 and 2y = 4x + 2 |
The equation of a circle
A circle is the set of all points whose distance from a fixed point (the centre) equals a given constant (the radius). From this definition, together with the distance formula, the standard equation of a circle follows directly.
The standard form: (x − a)² + (y − b)² = r², where (a, b) is the centre of the circle and r is the radius. Watch the signs: the centre is (a, b) with a minus in the equation, so if the equation is (x − 3)² + (y + 2)² = 25, the centre is (3, −2) — not (3, 2)! — and the radius is √25 = 5. At GCSE the circle is very often centred at the origin, x² + y² = r², but the general form does appear.
How to find the equation of a circle. You need two pieces of information: the centre and the radius. (a) If you are given the centre and a point on the circle — the radius is the distance between them. (b) If you are given the two ends of a diameter — the centre is the midpoint of the diameter, and the radius is half the length of the diameter. (c) If you are given the centre and a tangent — the radius is the distance from the centre to the tangent line.
A full example. Find the equation of the circle with centre (2, −1) passing through the point (5, 3). The radius is the distance from the centre to the point: r = √[(5 − 2)² + (3 − (−1))²] = √[9 + 16] = √25 = 5. So r² = 25, and the equation is (x − 2)² + (y + 1)² = 25.
The expanded (general) form: sometimes the circle is given as x² + y² + Dx + Ey + F = 0. To find the centre and radius you 'complete the square' on each pair. Example: x² + y² − 6x + 4y − 12 = 0 → (x² − 6x) + (y² + 4y) = 12 → (x − 3)² − 9 + (y + 2)² − 4 = 12 → (x − 3)² + (y + 2)² = 25. Centre (3, −2), radius 5. (Completing the square on a circle is strictly beyond the GCSE specification, but the technique is the same one you use for quadratics, and it turns up on some exam boards' harder questions.)
Where a line meets a circle
To find the points where a line meets a circle, solve the two equations together as simultaneous equations: make y (or x) the subject of the line's equation, substitute into the circle's equation, and you get a quadratic in one variable. The number of solutions of the quadratic decides the number of points of intersection.
| Discriminant of the quadratic | Geometric situation | Number of points |
|---|---|---|
| Δ > 0 | The line cuts the circle (a chord) | Two points |
| Δ = 0 | The line is a tangent to the circle | One point |
| Δ < 0 | The line does not meet the circle | No points |
Example. Find the points where the line y = x + 1 meets the circle x² + y² = 25. Substitute: x² + (x + 1)² = 25 → x² + x² + 2x + 1 = 25 → 2x² + 2x − 24 = 0 → x² + x − 12 = 0 → (x + 4)(x − 3) = 0. So x = −4 or x = 3. From the line: for x = −4 we get y = −3, and for x = 3 we get y = 4. The points of intersection: (−4, −3) and (3, 4). Two points — the line cuts the circle.
The tangent to a circle
A tangent is a line that touches the circle at exactly one point — the point of contact. The central property, on which almost every tangent question in the exam rests: the radius to the point of contact is perpendicular to the tangent. In other words the gradient of the tangent is the negative reciprocal of the gradient of the radius.
Finding the tangent at a given point on the circle — in three steps. (1) Work out the gradient of the radius from the centre to the point of contact. (2) The gradient of the tangent is its negative reciprocal (from m₁ × m₂ = −1). (3) Substitute the gradient and the point of contact into the point–gradient form.
A full example. You are given the circle (x − 1)² + (y − 2)² = 25, and the point of contact (4, 6) on it. The centre of the circle is (1, 2). The gradient of the radius from (1, 2) to (4, 6): m_radius = (6 − 2)/(4 − 1) = 4/3. The gradient of the tangent (the negative reciprocal): m_tangent = −3/4. The equation of the tangent through (4, 6): y − 6 = −3/4(x − 4), that is y = −3/4·x + 9. Check: the distance from the centre to this line is indeed 5 = the radius.
Another pattern — a tangent with a given gradient. If you are asked for the tangents to a circle with a fixed gradient m, write the family of lines y = mx + k, require the distance from the centre to the line to equal r, and solve for k. You usually get two values of k — two parallel tangents on opposite sides of the circle. (This variant is more of an A level question, but it shows how the tools fit together.)
Combined problems — line and circle together
The real exam question is almost always a chain: several parts, each relying on the one before. The secret is to identify which tool to pull out at each stage. Let's go through a typical example problem from start to finish.
The problem: a circle has centre O(3, 1) and passes through the point A(7, 4). (a) Find the equation of the circle. (b) Find the equation of the tangent to the circle at A. (c) The tangent crosses the x-axis at the point B — find B. (d) Work out the length of the segment OB.
- Part (a) — the radius is the distance OA: r = √[(7 − 3)² + (4 − 1)²] = √[16 + 9] = √25 = 5. The equation: (x − 3)² + (y − 1)² = 25.
- Part (b) — the gradient of the radius OA: (4 − 1)/(7 − 3) = 3/4. The gradient of the perpendicular tangent: −4/3. The equation through A(7, 4): y − 4 = −4/3(x − 7), that is y = −4/3·x + 40/3.
- Part (c) — crossing the x-axis means y = 0: 0 = −4/3·x + 40/3 → 4/3·x = 40/3 → x = 10. So B(10, 0).
- Part (d) — the length OB: √[(10 − 3)² + (0 − 1)²] = √[49 + 1] = √50 = 5√2.
Common mistakes
- The signs of the centre in the equation of the circle: in (x − a)² the centre is +a, and in (x + 3)² the centre is −3. Many pupils read the sign as it appears, and get it wrong.
- Confusing r with r²: the right-hand side of the circle's equation is r², not r. If r = 5, the right-hand side is 25. And if the right-hand side is 16, the radius is 4, not 16.
- A half-done perpendicular gradient: forgetting to flip the fraction or forgetting the sign. For 2/3 the perpendicular is −3/2, not −2/3 and not 3/2.
- Completing the square wrongly: when completing x² − 6x you need to add and subtract (6/2)² = 9, and not forget to balance both sides.
- Forgetting the second solution at an intersection: a quadratic with two roots gives two points of intersection. Stopping after one root = losing half the answer.
- Mixing up the variable: when substituting the line into the circle, substitute y (or x) everywhere it appears — not only in one of the terms.
A practice plan
Four focused days of practice are enough to close the topic to exam level, assuming you have the Year 10 foundations.
- Day 1 — foundations: 15 exercises on distance, midpoint, gradient and the equation of a line. Until each one is solved in one go without glancing at the formula.
- Day 2 — the circle: 10 exercises on finding the equation of a circle from different information (centre + point, diameter, general form with completing the square).
- Day 3 — intersection and tangent: 8 exercises on where a line meets a circle (including checking the discriminant) and 6 exercises on the tangent at a point and with a given gradient.
- Day 4 — full exam questions: 4 combined questions from past Higher papers under timed conditions, checking against the mark scheme only at the end.
Summary
Coordinate geometry at GCSE Higher is built on a small number of tools that come up again and again: distance, midpoint, gradient, the conditions m₁ = m₂ and m₁ × m₂ = −1, and the equation of the circle (x − a)² + (y − b)² = r². Every exam question — even the most combined and complex — is a combination of these tools in a certain order. The key is not to memorise whole solutions, but to recognise at each stage which tool is needed: a radius? Pull out distance. A tangent? Pull out perpendicularity. An intersection? Pull out simultaneous equations. A pupil who commands this chain will approach the question with confidence and complete it in full.
Frequently asked questions
What is the equation of a circle and how do you read the centre and radius from it?
The standard equation of a circle is (x − a)² + (y − b)² = r², where (a, b) is the centre and r the radius. Watch the reversed signs: in the equation (x − 3)² + (y + 2)² = 25 the centre is (3, −2) and the radius is √25 = 5. The right-hand side is always r² (the radius squared), not the radius itself. At GCSE the circle is often centred at the origin: x² + y² = r².
When are two lines parallel and when are they perpendicular?
Two lines are parallel when their gradients are equal: m₁ = m₂. They are perpendicular when the product of the gradients is minus one: m₁ × m₂ = −1, that is one gradient is the negative reciprocal of the other. For example, a line with gradient 2 is perpendicular to a line with gradient −1/2.
How do you find the tangent to a circle at a given point?
The tangent is perpendicular to the radius at the point of contact. So: (1) work out the gradient of the radius from the centre to the point of contact, (2) the gradient of the tangent is its negative reciprocal by m₁ × m₂ = −1, (3) substitute the gradient and the point of contact into the point–gradient form y − y₁ = m(x − x₁).
How do you know whether a line cuts a circle, touches it, or misses it?
Substitute the equation of the line into the equation of the circle and you get a quadratic. If the discriminant is positive (Δ > 0) — the line cuts the circle at two points; if it is zero (Δ = 0) — the line is a tangent (one point); if it is negative (Δ < 0) — the line does not meet the circle. An alternative route for tangency: the distance from the centre to the line equals the radius.
What is the difference between the coordinate geometry of Year 10 and Year 11?
In Year 10 the focus is on the foundations: points, distance, midpoint, gradient and the equation of a line (Foundation and Higher). In Year 11 on Higher the topic is extended to the circle — the equation of a circle, where a line meets a circle, and tangents — and above all to combined problems that join a line and a circle in the same diagram over several dependent parts.
A full paper in GCSE Higher style — including coordinate geometry, the circle and the tangent
Practise with a Higher mock paper ←