Circle Equations — Edexcel A-level Mathematics (Pure)

SPEC 3.2 - COORDINATE GEOMETRY

Equation of a Circle

Understanding and applying the equation of a circle in different forms

Learning Objectives

By the end of this lesson, you will be able to:

Understand the standard form of a circle equation: (x − a)² + (y − b)² = r²

Identify the centre and radius from the standard form

Recognise and work with the expanded form: x² + y² + 2gx + 2fy + c = 0

Convert between standard and expanded forms using completing the square

Determine whether a point lies inside, on, or outside a circle

Exam Focus

Circle equations appear frequently in A-Level exams, often combined with straight line questions (tangents, normals, intersections).

Circle from First Principles

Definition

A circle is the set of all points that are a fixed distance (the radius) from a fixed point (the centre).

If a circle has centre (a, b) and radius r, then any point (x, y) on the circle satisfies:

Distance from (x, y) to (a, b) = r

Using the distance formula:

Derivation

(x - a)^{2} + (y - b)^{2} = r

Squaring both sides gives the standard form:

(x - a)^{2} + (y - b)^{2} = r^{2}

Standard Form

(x - a)^{2} + (y - b)^{2} = r^{2}

Centre: (a, b) Radius: r

Watch the Signs!

The formula uses subtraction: (x − a) and (y − b)

If you see (x − 3)², then a = 3

If you see (x + 3)², this is (x − (−3))², so a = −3

If you see (y + 5)², this is (y − (−5))², so b = −5

Remember

The right-hand side is r², not r. Always square root to find the radius!

Any point (x, y) on the circle is distance r from centre (a, b)

Worked Example01

Write the equation of a circle with centre (3, −2) and radius 5.

1

Identify: Centre (a, b) = (3, −2), radius r = 5

2

Use standard form: (x − a)² + (y − b)² = r²

3

Substitute: (x − 3)² + (y − (−2))² = 5²

4

Simplify: (x − 3)² + (y + 2)² = 25

Answer: (x − 3)² + (y + 2)² = 25

Worked Example02

Find the centre and radius of the circle (x − 4)² + (y + 1)² = 16.

1

Compare with standard form: (x − a)² + (y − b)² = r²

2

From (x − 4)²: a = 4

3

From (y + 1)² = (y − (−1))²: b = −1

4

From r² = 16: r = √16 = 4

Answer: Centre (4, −1), Radius 4

Special Case: Centre at Origin

When the circle is centred at the origin (0, 0), the equation simplifies:

Circle Centred at Origin

x^{2} + y^{2} = r^{2}

Centre: (0, 0) Radius: r

Example 1

x^{2} + y^{2} = 9

Centre: (0, 0)

Radius: √9 = 3

Example 2

x^{2} + y^{2} = 25

Centre: (0, 0)

Radius: √25 = 5

Quick Check

If you see x² + y² = (number), the circle is centred at the origin.

Expanded Form

When we expand (x − a)² + (y − b)² = r², we get the expanded form (also called general form):

Expanded Form

x^{2} + y^{2} + 2 gx + 2 f y + c = 0

Relationship to Centre and Radius

PropertyFormulaCentre(−g, −f)Radius√(g² + f² − c)

Important

The coefficients are 2g and 2f, so divide by 2 to find g and f.

For a valid circle, we need g² + f² − c > 0

Completing the Square Review

To convert from expanded form to standard form, we use completing the square.

Method

For x² + bx:

x^{2} + b x = (x + \frac{b}{2})^{2} - (\frac{b}{2})^{2}

Example: x² + 6x

Half of 6 is 3

x² + 6x = (x + 3)² − 9

Example: y² − 4y

Half of −4 is −2

y² − 4y = (y − 2)² − 4

Strategy for Circles

Complete the square for x terms, then for y terms, then rearrange to get standard form.

Worked Example03

Convert x² + y² − 6x + 4y − 12 = 0 to standard form.

1

Group x and y terms: (x² − 6x) + (y² + 4y) = 12

2

Complete the square for x: x² − 6x = (x − 3)² − 9

3

Complete the square for y: y² + 4y = (y + 2)² − 4

4

Substitute: (x − 3)² − 9 + (y + 2)² − 4 = 12

5

Simplify: (x − 3)² + (y + 2)² = 12 + 9 + 4 = 25

Answer: (x − 3)² + (y + 2)² = 25

Worked Example04

Find the centre and radius of the circle x² + y² + 8x − 2y − 8 = 0.

1

Rearrange: (x² + 8x) + (y² − 2y) = 8

2

Complete square for x: x² + 8x = (x + 4)² − 16

3

Complete square for y: y² − 2y = (y − 1)² − 1

4

Substitute: (x + 4)² − 16 + (y − 1)² − 1 = 8

5

Simplify: (x + 4)² + (y − 1)² = 8 + 16 + 1 = 25

6

Read off: Centre (−4, 1), r² = 25 so r = 5

Answer: Centre (−4, 1), Radius 5

Points and Circles

To determine whether a point P(x₁, y₁) lies inside, on, or outside a circle:

Method

Calculate the distance from P to the centre, then compare with the radius r.

Test

For circle (x − a)² + (y − b)² = r², substitute P(x₁, y₁):If (x₁ − a)² + (y₁ − b)² ...Then P is...< r²Inside the circle= r²On the circle> r²Outside the circle

Worked Example05

Determine whether the point (5, 3) lies inside, on, or outside the circle (x − 2)² + (y − 1)² = 25.

1

Circle has centre (2, 1) and r² = 25

2

Substitute point (5, 3) into LHS of equation

3

(5 − 2)² + (3 − 1)² = 3² + 2² = 9 + 4 = 13

4

Compare: 13 < 25 (i.e., result < r²)

Answer: The point (5, 3) lies INSIDE the circle

Worked Example06

A circle has centre (3, 4) and passes through the origin. Find the equation of the circle.

1

Centre is (3, 4) and the circle passes through (0, 0)

2

The radius = distance from centre to origin

3

r = √[(3 − 0)² + (4 − 0)²] = √[9 + 16] = √25 = 5

4

Write equation: (x − 3)² + (y − 4)² = 25

5

Check: (0 − 3)² + (0 − 4)² = 9 + 16 = 25 ✓

Answer: (x − 3)² + (y − 4)² = 25

Practice

Question 1: Write the equation of the circle with centre (−2, 5) and radius 7.

1

Centre (a, b) = (−2, 5), radius r = 7

2

(x − (−2))² + (y − 5)² = 7²

3

(x + 2)² + (y − 5)² = 49

Answer: (x + 2)² + (y − 5)² = 49

Practice

Question 2: Find the centre and radius of the circle x² + y² − 10x + 6y + 9 = 0.

1

Rearrange: (x² − 10x) + (y² + 6y) = −9

2

Complete square for x: (x − 5)² − 25

3

Complete square for y: (y + 3)² − 9

4

(x − 5)² − 25 + (y + 3)² − 9 = −9

5

(x − 5)² + (y + 3)² = −9 + 25 + 9 = 25

Answer: Centre (5, −3), Radius 5

Practice

Question 3: Determine whether the point (7, 1) lies inside, on, or outside the circle (x − 3)² + (y + 2)² = 36.

1

Circle: centre (3, −2), r² = 36

2

Substitute (7, 1): (7 − 3)² + (1 − (−2))²

3

= 4² + 3² = 16 + 9 = 25

4

Compare: 25 < 36

Answer: The point (7, 1) lies INSIDE the circle

Summary

Key Formulas

FormEquationCentreRadius

Standard(x − a)² + (y − b)² = r²(a, b)r

Originx² + y² = r²(0, 0)r

Expandedx² + y² + 2gx + 2fy + c = 0(−g, −f)√(g² + f² − c)

Exam Tips

Watch the signs when reading centre from standard form

Remember to square root r² to find radius

When completing the square, add the same value to both sides

To test a point: substitute and compare with r²