A-Level · Mathematics · AQA · Mark scheme decoded
AQA A-Level Mathematics: Circle Coordinate Geometry — mark scheme explained
The short answer
In this section, we will explore the coordinate geometry of circles. This includes understanding and using the equation of a circle in its standard form, completing the square to find the centre and radius, and applying key properties of circles.
The question
Find the equation of a circle with centre (3, -4) and radius 5.
[Paraphrased for study — not reproduced from any exam paper.]
Mark scheme, decoded
How the examiner actually awards the marks on this topic.
Gradora's own decode of the marking approach — not the exam board's published mark scheme.
How marks are awarded
For questions involving circle coordinate geometry, marks are typically awarded for correctly completing the square, identifying the centre and radius, applying key properties, and providing clear and accurate working. Ensure your steps are logical and well-organized.
What the command words demand
- Find
- Calculate or determine the value of a specific quantity.
- Show
- Demonstrate a result using given information and appropriate methods.
- Prove
- Provide a logical argument to establish a mathematical statement.
- Determine
- Identify or calculate a required value or property.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate about 5 minutes per mark to ensure you have enough time to complete each step accurately and check your work.
- The standard form of the equation of a circle is (x - a) 2 + (y - b) 2 = r 2 .0 marks
- Here, the centre (a, b) is (3, -4) and the radius r is 5.1 mark
- Substitute these values into the equation: (x - 3) 2 + (y + 4) 2 = 5 2 .1 mark
- Simplify the right-hand side: (x - 3) 2 + (y + 4) 2 = 25.1 mark
Final answer: (x - 3) 2 + (y + 4) 2 = 25
Work through every step correctly and you earn all 3 marks.
Another worked example
Find the centre and radius of the circle given by the equation x 2 + y 2 - 6x + 8y + 9 = 0.
- To find the centre and radius, we need to complete the square for both x and y terms.0 marks
- For x: x 2 - 6x can be written as (x - 3) 2 - 9.1 mark
- For y: y 2 + 8y can be written as (y + 4) 2 - 16.1 mark
- Substitute these into the equation: (x - 3) 2 - 9 + (y + 4) 2 - 16 + 9 = 0.1 mark
- Simplify: (x - 3) 2 + (y + 4) 2 - 16 = 0.0 marks
- Add 16 to both sides: (x - 3) 2 + (y + 4) 2 = 16.1 mark
- The centre is (3, -4) and the radius is √16 = 4.1 mark
Final answer: Centre: (3, -4), Radius: 4
Work through every step correctly and you earn all 5 marks.
Common mistakes
Forgetting to complete the square when finding the centre and radius.
Why it happens: Students often try to directly read off the centre and radius from a given equation without transforming it into the standard form.
Fix: Always rewrite the equation in the form (x - a) 2 + (y - b) 2 = r 2 by completing the square for both x and y terms.
Incorrectly calculating the radius when completing the square.
Why it happens: Students may forget to take the square root of the constant term after completing the square, leading to an incorrect radius.
Fix: After completing the square, ensure you take the square root of the constant term on the right-hand side to find the radius.
Misapplying the property that the angle in a semicircle is a right angle.
Why it happens: Students sometimes apply this property incorrectly, especially when the triangle is not clearly inscribed in a semicircle.
Fix: Always verify that the triangle is inscribed in a semicircle and that the angle at the circumference is opposite the diameter.
Forgetting that the perpendicular from the centre to a chord bisects the chord.
Why it happens: Students may overlook this property when solving problems involving chords and their lengths.
Fix: Remember that the perpendicular from the centre to a chord always bisects the chord. Use this property to find the length of the chord or its midpoint.
Incorrectly identifying the relationship between the radius and the tangent.
Why it happens: Students may confuse the perpendicularity condition, leading to incorrect angles or distances in problems involving tangents.
Fix: Always remember that the radius at any point on the circumference is perpendicular to the tangent at that point. Use this property to find angles or distances.
Failing to check if a given equation represents a circle.
Why it happens: Students sometimes assume any quadratic equation in x and y represents a circle without verifying the form.
Fix: Always verify that the equation can be written in the standard form (x - a) 2 + (y - b) 2 = r 2 . If it cannot, the given equation may represent another conic section.
Incorrectly calculating the midpoint of a chord.
Why it happens: Students may make arithmetic errors when using the midpoint formula or misinterpret the coordinates of the endpoints.
Fix: Double-check your calculations and ensure you correctly apply the midpoint formula: (x 1 + x 2 ) / 2, (y 1 + y 2 ) / 2.
Where the marks go
The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.
| Question type | What you’re asked to do | Marks |
|---|---|---|
| Equation Of Circle | Write the equation of a circle given its centre and radius. | 3 |
| Circle Equation Analysis | Complete the square to find a circle's centre and radius from its general equation. | 5 |
| Circle Equation From Diameter | Find a circle's equation given the coordinates of two endpoints of its diameter. | 6 |
| Circle Chord Length | Use the perpendicular distance from the centre to a chord to find its length. | 6 |
| Total across these question types | 20 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.