What mistakes do students make with Coordinate geometry in the (x, y) plane?

Reading the centre as (a, b) without reversing signs. (x − 3)² gives x-coordinate 3. Forgetting to subtract the constant when completing the square. Balance both sides so the equation becomes (x − a)² + (y − b)² = r². Using the radius gradient itself as the tangent gradient. Take the negative reciprocal, since radius and tangent are perpendicular.

What are the exam tips for Coordinate geometry in the (x, y) plane in Mathematics Pearson Edexcel A-Level?

Sketch the circle with its centre marked to avoid sign errors. When a chord or tangent is involved, draw the radius to the relevant point and mark the right angle. Check the radius by substituting a known point on the circle into the equation.

Pearson Edexcel · A-Level · Mathematics

Course version: 9MA0 · Version from Mar 2026

Understand and use the coordinate geometry of the circle including using…

Full specification wording

Coordinate geometry in the (x, y) plane: 3.2 Understand and use the coordinate geometry of the circle including using the equation of a circle in the form (x − a)² + (y − b)² = r²; completing the square to find the centre and radius of a circle; use of the following properties: the angle in a semicircle is a right angle; the perpendicular from the centre to a chord bisects the chord; the radius of a circle at a given point on its circumference is perpendicular to the tangent to the circle at that point.

A circle with centre (a, b) and radius r has equation (x − a)² + (y − b)² = r², so (x − 3)² + (y + 1)² = 25 has centre (3, −1) and radius 5.

Checked against the Pearson Edexcel 9MA0 specification (opens in a new tab) on 25 September 2026.

Point 2 of 4
4 key terms

Study notes

From the checked teaching notes for this specification point.

The idea

What earns the marks

Reads the centre and radius directly from (x − a)² + (y − b)² = r², taking care with the signs of a and b.
Completes the square in both x and y to convert a general circle equation into centre-radius form.
Uses the angle in a semicircle being a right angle to set up a perpendicular or right-angled triangle condition.
Applies the perpendicular from the centre to a chord bisecting the chord to find a midpoint or a length.
Uses the radius being perpendicular to the tangent to find a tangent gradient as the negative reciprocal of the radius gradient.

Common mistakes and how to put them right

Reading the centre as (a, b) without reversing signs.
Put it right: (x − 3)² gives x-coordinate 3.
Forgetting to subtract the constant when completing the square.
Put it right: Balance both sides so the equation becomes (x − a)² + (y − b)² = r².
Using the radius gradient itself as the tangent gradient.
Put it right: Take the negative reciprocal, since radius and tangent are perpendicular.

Exam tips

Sketch the circle with its centre marked to avoid sign errors.
When a chord or tangent is involved, draw the radius to the relevant point and mark the right angle.
Check the radius by substituting a known point on the circle into the equation.

Try the lesson

The start of the Coordinate geometry in the (x, y) plane: 3.2 Understand and use the coordinate geometry of the circle including using the equation of a circle in the form (x − a)² + (y − b)² = r²; completing the square to find the centre and radius of a circle; use of the following properties: the angle in a semicircle is a right angle; the perpendicular from the centre to a chord bisects the chord; the radius of a circle at a given point on its circumference is perpendicular to the tangent to the circle at that point. lesson, made from checked teaching notes for this specification point.

Have a go first

A circle has equation (x − 4)² + (y + 3)² = 25. State the centre and radius of the circle.

Tip: press 1 to 4 to choose

Listen

A circle with centre (a, b) and radius r has equation (x − a)² + (y − b)² = r², so (x − 3)² + (y + 1)² = 25 has centre (3, −1) and radius 5. If the equation is expanded, such as x² + y² − 6x + 2y − 15 = 0, complete the square in x and in y to recover the centre and radius. Three circle theorems support coordinate work: the angle in a semicircle is a right angle, so a triangle with a diameter as one side has a right angle at the circumference; the perpendicular from the centre to a chord bisects the chord; and the radius at a point on the circumference is perpendicular to the tangent there. These let you find centres, chords and tangent equations. A strong answer earns marks for these points. Reads the centre and radius directly from (x − a)² + (y − b)² = r², taking care with the signs of a and b. Completes the square in both x and y to convert a general circle equation into centre-radius form. Uses the angle in a semicircle being a right angle to set up a perpendicular or right-angled triangle condition. Applies the perpendicular from the centre to a chord bisecting the chord to find a midpoint or a length. A common slip: Reading the centre as (a, b) without reversing signs: the error is taking (x − 3)² as centre (−3, b). Instead, (x − 3)² gives x-coordinate 3. Exam tip: Sketch the circle with its centre marked to avoid sign errors.

See it

A top-down flowchart. The top node is the circle equation (x − a)² + (y − b)² = r². It leads to two nodes: reading the centre (a, b) and radius r, and the expanded equation. The expanded equation leads to completing the square in x and in y, which leads back to reading the centre and radius. From reading the centre and radius, three branches lead to the three circle theorems: the angle in a semicircle is a right angle; the perpendicular from the centre to a chord bisects the chord; and the radius at a point is perpendicular to the tangent. All three theorem nodes lead to a final node: find centres, chords and tangent equations. — It shows how the centre-radius form and completing the square give the centre and radius, and how the three circle theorems support finding centres, chords and tangent equations.

Key words

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Centre-radius form

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Completing the square

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Angle in a semicircle

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Perpendicular from the centre to a chord

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See every definition

Centre-radius form: The equation of a circle written as (x − a)² + (y − b)² = r², giving centre (a, b) and radius r.
Completing the square: Used in x and in y to convert a general circle equation into centre-radius form, recovering the centre and radius.
Angle in a semicircle: A right angle, so a triangle with a diameter as one side has a right angle at the circumference.
Perpendicular from the centre to a chord: It bisects the chord, which lets you find a midpoint or a length.

Label the diagram

Some boxes have lost their names. Pick the right name for each number.

The diagram again, with 3 boxes numbered instead of named. A top-down flowchart. The top node is the circle equation (x − a)² + (y − b)² = r². It leads to two nodes: reading the centre (a, b) and radius r, and the expanded equation. The expanded equation leads to completing the square in x and in y, which leads back to reading the centre and radius. From reading the centre and radius, three branches lead to the three circle theorems: the angle in a semicircle is a right angle; the perpendicular from the centre to a chord bisects the chord; and the radius at a point is perpendicular to the tangent. All three theorem nodes lead to a final node: find centres, chords and tangent equations.

1Box 1
2Box 2
3Box 3

0 of 3 labelled

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Test your understanding of Coordinate geometry in the (x, y) plane: 3.2 Understand and use the coordinate geometry of the circle including using the equation of a circle in the form (x − a)² + (y − b)² = r²; completing the square to find the centre and radius of a circle; use of the following properties: the angle in a semicircle is a right angle; the perpendicular from the centre to a chord bisects the chord; the radius of a circle at a given point on its circumference is perpendicular to the tangent to the circle at that point.

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Understand and use the coordinate geometry of the circle including using…