A16: Circles
Recognise and use the equation of a circle with centre at the origin; find equation of tangent at a point
📋 Key Concepts
Equation of a circle: A circle with centre (0, 0) and radius r has equation:
x² + y² = r²
General form: For a circle with centre (a, b) and radius r:
(x - a)² + (y - b)² = r²
📝 Equation of a Circle (Centre at Origin)
Example 1
Write the equation of a circle with centre (0, 0) and radius 5.
Solution:
x² + y² = 5²
x² + y² = 25
Example 2
Find the radius of the circle x² + y² = 16
Solution:
x² + y² = r² = 16
r² = 16
r = 4
Example 3
Does the point (3, 4) lie on the circle x² + y² = 25?
Solution:
Substitute: 3² + 4² = 9 + 16 = 25 ✓
Yes, the point (3, 4) lies on the circle.
📝 Finding Points on a Circle
Example 4
A circle has equation x² + y² = 100. Find the y-coordinate when x = 6.
Solution:
Substitute x = 6:
6² + y² = 100
36 + y² = 100
y² = 64
y = ±8
Points: (6, 8) and (6, -8)
📝 Equation of a Circle (Any Centre)
Example 5
Write the equation of a circle with centre (2, -3) and radius 4.
Solution:
Using (x - a)² + (y - b)² = r²:
(x - 2)² + (y - (-3))² = 4²
(x - 2)² + (y + 3)² = 16
Example 6
Find the centre and radius of (x + 1)² + (y - 4)² = 9
Solution:
Compare with (x - a)² + (y - b)² = r²
a = -1, b = 4, r² = 9
Centre: (-1, 4), Radius: 3
📝 Finding the Tangent to a Circle
Key property: The tangent at any point on a circle is perpendicular to the radius at that point.
Method:
- Find the gradient of the radius (from centre to point)
- Find the negative reciprocal (perpendicular gradient)
- Use y - y₁ = m(x - x₁) to find the tangent equation
Example 7
Find the equation of the tangent to x² + y² = 25 at the point (3, 4).
Solution:
Step 1: Gradient of radius from (0,0) to (3,4):
m_radius = 4 - 0⁄3 - 0 = 4⁄3
Step 2: Gradient of tangent (perpendicular):
m_tangent = -3⁄4
Step 3: Use point (3, 4) and gradient -3⁄4:
y - 4 = -3⁄4(x - 3)
4(y - 4) = -3(x - 3)
4y - 16 = -3x + 9
3x + 4y = 25
📝 Expanding Circle Equations
Example 8
Expand (x - 3)² + (y + 2)² = 16
Solution:
x² - 6x + 9 + y² + 4y + 4 = 16
x² + y² - 6x + 4y + 13 = 16
x² + y² - 6x + 4y - 3 = 0
Example 9
Find the centre and radius of x² + y² + 6x - 4y - 12 = 0
Solution:
Complete the square for x and y:
(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
🧮 Worked Examples — Exam Style
Circle equations: centre origin → x² + y² = r². Any centre → (x − a)² + (y − b)² = r². The MINUS signs in the brackets match the centre coordinates.
Write the equation of the circle centred at (0, 0) with radius 9.
x² + y² = 81
B1 for 81 (radius SQUARED). Writing 9 on the right is the standard error.
Does the point (3, 4) lie on x² + y² = 30?
3² + 4² = 9 + 16 = 25 ≠ 30 → no — it lies INSIDE the circle
M1 for the substitution, A1 with the inside/outside interpretation. Less than r² → inside; greater → outside.
Exam wording to expect: "write down the equation of the circle", "find the centre and radius", "does the point lie on the circle?".
❓ Practice Questions
Q1: Write the equation of a circle with centre (0, 0) and radius 7.
Q2: Find the radius of x² + y² = 36.
Q3: Does (5, 12) lie on x² + y² = 169?
Q4: Find the centre and radius of (x - 4)² + (y + 1)² = 9.
Q5: Find the gradient of the tangent to x² + y² = 25 at point (4, 3).
Q6: Find the centre of x² + y² + 8x - 2y - 8 = 0.
Q6: Find the equation of the circle with centre (2, −3) and radius 5
Q7: Find the centre and radius of (x + 5)² + (y − 2)² = 20
✅ Answers
- x² + y² = 49
- 6
- Yes: 5² + 12² = 25 + 144 = 169 ✓
- Centre (4, -1), Radius 3
- m_radius = 3⁄4, so m_tangent = -4⁄3
- Centre (-4, 1)
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. x² + y² = 49
Q2. r = √36 = 6
Q3. 25 + 144 = 169 ✓ → yes, on the circle
Q4. Centre (4, −1), radius 3
Q5. Radius gradient = 3/4 → tangent gradient = −4/3 (perpendicular)
Q6. Complete the square: (x + 4)² + (y − 1)² = 8 + 16 + 1 = 25 → centre (−4, 1), radius 5
Q7. Centre (−5, 2), radius √20 = 2√5 ≈ 4.47
📌 Key Facts to Memorise
- Centre origin: x² + y² = r² · any centre (a, b): (x − a)² + (y − b)² = r²
- The bracket signs FLIP the centre signs: (x + 5)² means a = −5
- Radius = √(right-hand side) — the number is r², not r
- Point on circle: substitute; it must satisfy the equation exactly
- Tangent ⊥ radius at the point of contact (perpendicular gradients)
🗺️ In the Exam
Circle equations: Higher-tier content in this folder — 3–4 marks on Higher papers, including completing the square to find the centre.
Links to: coordinate geometry (G11), completing the square (A18), Pythagoras (G20).
🧭 Board Focus — AQA
Spec 8300. Three papers, each 80 marks and 1h30: Paper 1 non-calculator, Papers 2 and 3 calculator. Foundation/Higher tiers. Roughly a third of the marks sit under AO2 (reasoning) and AO3 (problem-solving), so show every line of working — method marks survive arithmetic slips.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- x² + y² = r² for circle at origin
- (x - a)² + (y - b)² = r² for centre (a, b)
- Tangent is perpendicular to radius
- Use negative reciprocal for perpendicular gradient
- Complete the square to find centre and radius from expanded form
🧠 Problem-Solving Strategies
Problem-Solving
To find a tangent: find the gradient of the radius (centre to point), take the negative reciprocal for the perpendicular gradient, then use y - y1 = m(x - x1). To find centre/radius from expanded form, complete the square for both x and y.
Multi-Step Problem
A circle has equation x² + y² = 34. Point P has coordinates (3, 5). (a) Verify P lies on the circle. (b) Find the equation of the tangent at P. (c) Find where the tangent crosses the y-axis.
Solution:
(a) 3² + 5² = 9 + 25 = 34 ✓
(b) Gradient of radius = 5/3. Tangent gradient = -3/5. y - 5 = -3/5(x - 3). 5y - 25 = -3x + 9. 3x + 5y = 34
(c) When x = 0: 5y = 34, y = 34/5 = 6.8. Point: (0, 6.8)
⚠️ Common Errors
Watch Out!
1. Wrong: (x + 3)² + (y - 2)² = 25 has centre (3, -2) Correct: Centre is (-3, 2) — the signs in the equation are opposite to the coordinates
2. Wrong: The tangent has the same gradient as the radius Correct: The tangent is perpendicular to the radius — use the negative reciprocal
3. Wrong: x² + y² - 6x = 0 has radius 6 Correct: Complete the square: (x-3)² + y² = 9, so radius = 3
✍️ 6-Mark Exam Question
Extended Answer
6 marks: (a) Find the centre and radius of the circle x² + y² + 4x - 6y - 3 = 0. (b) Does (1, 4) lie inside, on or outside the circle? (c) Find the equation of the tangent at point (2, 3+√3) on the circle.
(a) (x² + 4x) + (y² - 6y) = 3. (x + 2)² - 4 + (y - 3)² - 9 = 3. (x + 2)² + (y - 3)² = 16. Centre (-2, 3), radius 4.
(b) Distance from (1, 4) to (-2, 3) = √(9 + 1) = √10 ≈ 3.16. Since 3.16 < 4, point is inside the circle.
(c) Gradient from centre to point: (3+√3 - 3)/(2-(-2)) = √3/4. Tangent gradient = -4/√3. Equation: y - (3+√3) = -4/√3(x - 2).
Mark scheme: (a) 2 marks. (b) 2 marks for distance and conclusion. (c) 2 marks.
📊 AO3: Reason & Interpret
Reasoning and Interpretation
A circular fountain has equation x² + y² = 100 (units in metres). A path has equation 3x + 4y = 50.
(a) Find the shortest distance from the centre to the path.
(b) Does the path cross the fountain?
(c) If the fountain radius increases by 2m, will the path now cross it?
Answers: (a) Distance from (0,0) to line = |50|/√(9+16) = 50/5 = 10 m. (b) No — the distance equals the radius exactly, so the path is a tangent. (c) New radius = 12 m > 10 m, so yes, the path would now cross the fountain at two points.
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