AQA GCSE Mathematics Paper 2 (Higher), June 2025: Question 21
5 marks · Hard difficulty · Multi-step Problem
Find the equation of a circle centered at the origin with radius √45, and find the equation of the tangent to the circle at the point (6, -3).
Practise this questionQuestion
Mark scheme
Show the mark scheme
How to answer it
Circle Equations & Tangents to a Circle
What this question tests
This Higher tier question assesses your mastery of geometric coordinate geometry, specifically:
- Recalling and applying the equation of a circle centred at the origin: x² + y² = r² .
- Finding the gradient of a radius using coordinates of the centre (0, 0) and a point on the circumference.
- Applying the circle theorem that a tangent is perpendicular to the radius at the point of contact ( m₁ × m₂ = -1 ).
- Finding the straight-line equation of a tangent in the form y = mx + c .
Writing the Equation of the Circle
A circle with centre O and radius √45
✅ Correct Answer
x² + y² = 45
Also accepted: (x - 0)² + (y - 0)² = 45 or x² + y² = (√45)² .
💡 Key Knowledge
- A circle centred at (0, 0) has the standard equation:
x² + y² = r² - Here, r = √45 , so r² = (√45)² = 45 .
❌ Common Errors
- Writing x² + y² = √45 (forgetting to square the radius).
- Using unrelated variables such as a² + b² = 45 (loses the mark).
- Squaring incorrectly, e.g. writing 2025 instead of 45 .
🧠 Exam Technique
This is a quick 1-mark recall question. Ensure your answer is written as an equation involving both x and y . Do not leave it as an expression like x² + y² .
• B1: Fully correct equation ( x² + y² = 45 oe).
Finding the Equation of the Tangent
Work out the equation of the tangent at point P(6, -3) in the form y = mx + c
📐 Step-by-Step Calculation
1 Find the gradient of radius OP:
Centre is (0, 0) and point is P(6, -3) .
Gradient of radius = (y₂ - y₁) / (x₂ - x₁) = (-3 - 0) / (6 - 0) = -3 / 6 = -1/2
2 Find the gradient of the perpendicular tangent:
Tangent is perpendicular to the radius: m_tangent = -1 / m_radius
m = -1 ÷ (-1/2) = 2
3 Find the y-intercept (c) using point P(6, -3):
Substitute m = 2 , x = 6 , and y = -3 into y = mx + c :
-3 = 2(6) + c
-3 = 12 + c
c = -3 - 12 = -15
4 State the final equation:
y = 2x - 15
✅ Final Answer
y = 2x - 15
Acceptable variants: y = -15 + 2x or unsimplified fractions such as y = 6/3 x - 15 .
💡 Key Knowledge
- Perpendicular rule: If gradient of radius is m , the tangent gradient is the negative reciprocal -1/m .
- Alternative line formula: y - y₁ = m(x - x₁)
y - (-3) = 2(x - 6)
y + 3 = 2x - 12 → y = 2x - 15
❌ Common Errors & Traps
- Sign slip on gradient: Writing (-3)/6 = 1/2 leading to a tangent gradient of -2 . Looking at the diagram, the tangent slopes upwards, so its gradient must be positive!
- Using √45 as the gradient: Confusing the length of the radius with the slope of the radius.
- Forgetting negative reciprocal: Using m = -1/2 as the tangent's gradient instead of inverting and changing sign.
- Stopping early: Finding c = -15 but not writing the complete equation y = 2x - 15 .
🧠 Mark Scheme & Examiner Insight
• M1: Method to find gradient of OP : (-3 - 0) / (6 - 0) or -1/2 .
• M1: Negative reciprocal method: -1 ÷ (-1/2) or stating gradient = 2 .
• M1dep: Dependent on 2nd M1. Substituting (6, -3) with their gradient into a valid straight line equation.
• A1: Fully correct equation: y = 2x - 15 .
Tip: Even if you make an arithmetic error finding the radius gradient, showing clear negative reciprocal method allows you to collect method follow-through marks!
Topics
Algebra · Geometry and measures · 3.2.2 Graphs · 3.4.1 Properties and constructions
Question and mark scheme from the AQA GCSE Mathematics examination, Paper 2 (Higher), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.