Equation of a Circle and Tangents
On GCSE Higher, every circle you meet in algebra is centred at the origin, so the equation is always x² + y² = r². Questions ask for the radius, intersections with a line, and the equation of a tangent. The tangent question is a classic grade 8–9 problem.
1The equation x² + y² = r²
A point (x, y) is on a circle of radius r centred at the origin when its distance from (0, 0) is r. By Pythagoras, that means x² + y² = r².
| Equation | Radius |
|---|---|
| x² + y² = 49 | 7 |
| x² + y² = 20 | √20 = 2√5 |
| x² + y² = 6.25 | 2.5 |
Common slip
- Saying the radius of x² + y² = 49 is 49. The number on the right is r², so take the square root.
2Is a point on, inside or outside the circle?
Substitute the point into x² + y² and compare with r²: equal means on the circle, smaller means inside, bigger means outside.
Worked example
Is (6, 7) inside or outside x² + y² = 81?
- 6² + 7² = 36 + 49 = 85.
- 85 > 81.
3Where a line meets a circle
Substitute the line into the circle equation to get a quadratic in x, solve it, then find each y.
Worked example
Find where y = x + 1 meets x² + y² = 25.
- x² + (x + 1)² = 25 → 2x² + 2x + 1 = 25 → x² + x − 12 = 0.
- (x + 4)(x − 3) = 0 → x = −4 or x = 3.
- y = x + 1 gives y = −3 and y = 4.
4The tangent at a point
A tangent is perpendicular to the radius at the point of contact. So:
- Find the gradient of the radius from (0, 0) to the point.
- The tangent’s gradient is the negative reciprocal.
- Use y − y₁ = m(x − x₁).
Worked example
Find the equation of the tangent to x² + y² = 25 at (3, 4).
- Radius gradient = 43.
- Tangent gradient = −34.
- y − 4 = −34(x − 3) → 4y − 16 = −3x + 9.
Exam tipThe negative reciprocal of ab is −ba. Check: the two gradients multiply to −1.
Common mistakes
- Giving r² as the radius.
- Using the gradient of the radius as the gradient of the tangent.
- Forgetting to find both y-values when a line meets a circle.
- Expanding (x + 1)² as x² + 1.
★Key Vocabulary Recap
- Radius
- Distance from the centre to the circle.
- Tangent
- A line that touches a circle at one point.
- Negative reciprocal
- For perpendicular lines, m₁ × m₂ = −1.
- Origin
- The point (0, 0).
Specification reference (Pearson Edexcel GCSE (9–1) Mathematics 1MA1: A16 (and A19 for line–circle intersections))
- A16: recognise and use the equation of a circle with centre at the origin; find the equation of a tangent to a circle at a given point.
- A19: solve two simultaneous equations in two variables (linear/quadratic) algebraically; find approximate solutions using a graph.
Next stepReady to practise? Try the 8 exam-style questions, then check your working against the worked solutions.
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Practice questions
Equation of a Circle and Tangents — Practice Questions
8 exam-style questions (26 marks), building from warm-up to grade 8–9. Show full working: Edexcel gives method marks even when the final answer is wrong.
Higher · Grades 7–9
- (1 mark)Grade 7Write down the radius of the circle x² + y² = 49.
- (2 marks)Grade 7Show that the point (5, −12) lies on the circle x² + y² = 169.
- (2 marks)Grade 7A circle has centre (0, 0) and passes through (−6, 8). Find its equation.
- (5 marks)Grade 8Solve algebraically the simultaneous equations x² + y² = 25 and y = 7 − x.
- (3 marks)Grade 8Find the equation of the tangent to x² + y² = 25 at the point (3, 4). Give your answer in the form ax + by = c.
- (4 marks)Grade 8The line y = 2x meets the circle x² + y² = 20 at two points. Find their coordinates.
- (4 marks)Grade 9The tangent to x² + y² = 20 at the point (2, 4) meets the x-axis at A. Find the coordinates of A.
- (5 marks)Grade 9P(−6, 8) lies on x² + y² = 100. The tangent at P meets the x-axis at A and the y-axis at B. Find the area of triangle AOB, where O is the origin.
Hint
Find the tangent’s equation first, then where it crosses each axis.
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Worked solutions
Show worked solutions
Equation of a Circle and Tangents — Worked Solutions
Full worked solutions to all 8 questions (26 marks), set out the way an Edexcel examiner wants to see them.
Higher · Grades 7–9
- (1 mark)Grade 7Write down the radius of the circle x² + y² = 49.Solution
- r² = 49, so r = 7.
7 - (2 marks)Grade 7Show that the point (5, −12) lies on the circle x² + y² = 169.Solution
- 5² + (−12)² = 25 + 144 = 169 ✓
Shown - (2 marks)Grade 7A circle has centre (0, 0) and passes through (−6, 8). Find its equation.Solution
- r² = (−6)² + 8² = 100.
x² + y² = 100 - (5 marks)Grade 8Solve algebraically the simultaneous equations x² + y² = 25 and y = 7 − x.Solution
- x² + (7 − x)² = 25 → 2x² − 14x + 49 = 25 → 2x² − 14x + 24 = 0.
- x² − 7x + 12 = 0 → (x − 3)(x − 4) = 0.
- x = 3, y = 4; x = 4, y = 3.
(3, 4) and (4, 3) - (3 marks)Grade 8Find the equation of the tangent to x² + y² = 25 at the point (3, 4). Give your answer in the form ax + by = c.Solution
- Radius gradient 4/3, so tangent gradient −3/4.
- y − 4 = −¾(x − 3) → 4y − 16 = −3x + 9.
3x + 4y = 25 - (4 marks)Grade 8The line y = 2x meets the circle x² + y² = 20 at two points. Find their coordinates.Solution
- x² + 4x² = 20 → 5x² = 20 → x² = 4.
- x = 2, y = 4 or x = −2, y = −4.
(2, 4) and (−2, −4) - (4 marks)Grade 9The tangent to x² + y² = 20 at the point (2, 4) meets the x-axis at A. Find the coordinates of A.Solution
- Radius gradient 4/2 = 2, so tangent gradient −½.
- y − 4 = −½(x − 2) → y = −½x + 5.
- At y = 0: x = 10.
A = (10, 0) - (5 marks)Grade 9P(−6, 8) lies on x² + y² = 100. The tangent at P meets the x-axis at A and the y-axis at B. Find the area of triangle AOB, where O is the origin.Solution
- Radius gradient 8/(−6) = −4/3, so tangent gradient 3/4.
- y − 8 = ¾(x + 6) → y = ¾x + 12.5.
- B = (0, 12.5). At y = 0: x = −50/3, so A = (−50/3, 0).
- Area = ½ × 503 × 12.5 = 6256 ≈ 104.2.
625/6 ≈ 104 square units
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