Chapter 06 — Coordinate Geometry | Part 2: Circles, Equations, Centre and Radius

MATHEMATICS HIPHOP ACADEMY

CHAPTER 06

COORDINATE GEOMETRY

PART 2: CIRCLES, EQUATIONS, CENTRE AND RADIUS

A-Level Pure Mathematics • Detailed Notes • Worked Examples • Exercises

Learning Objectives

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

• Define a circle using its centre and radius.
• Derive the Cartesian equation of a circle using the distance formula.
• Write the equation of a circle given its centre and radius.
• Identify the centre and radius from the standard equation.
• Convert an expanded circle equation into standard form by completing the square.
• Find the equation of a circle when the endpoints of a diameter are given.
• Find the equation of a circle passing through given points.
• Determine whether a given point lies inside, on, or outside a circle.
• Use circle equations to solve coordinate geometry problems.

1. Introduction to Circles in Coordinate Geometry

A circle is the locus of a point that moves in a plane while remaining at a constant distance from a fixed point.

The fixed point is called the centre, while the constant distance is called the radius.

For a circle with centre C(a, b) and radius r, every point P(x, y) on its circumference is exactly r units from the centre.

A circle with centre C and radius r. Every point on its circumference is the same distance from C.

1.1 Important Parts of a Circle

• Centre: the fixed point at the middle of the circle.
• Radius: the distance from the centre to the circumference.
• Diameter: a chord passing through the centre. Its length is twice the radius.
• Circumference: the boundary or complete curved edge of the circle.
• Chord: a straight line segment joining two points on the circumference.

Remember:

Diameter = 2 × Radius

2. Deriving the Equation of a Circle

We can derive the equation of a circle using the distance formula studied in Part 1.

Consider a circle with:

• Centre C(a, b)
• Radius r
• Any point P(x, y) on its circumference

Since P lies on the circumference, the distance CP is equal to the radius r.

Using the distance formula:

CP = √[(x − a)² + (y − b)²]

But CP = r. Therefore:

√[(x − a)² + (y − b)²] = r

Squaring both sides gives:

STANDARD EQUATION OF A CIRCLE

(x − a)² + (y − b)² = r²

Centre = (a, b); radius = r.

This is the standard equation of a circle with centre (a, b) and radius r.

2.1 Circle with Centre at the Origin

If the centre is at the origin O(0, 0), then a = 0 and b = 0.

Substituting into the standard equation:

(x − 0)² + (y − 0)² = r²

CIRCLE CENTRED AT THE ORIGIN

x² + y² = r²

Worked Example 1

Find the equation of a circle with centre at the origin and radius 5 units.

Solution

Using x² + y² = r²:

r = 5, so r² = 25.

Equation: x² + y² = 25.

Worked Example 2

Find the equation of a circle with centre (3, −2) and radius 4 units.

Solution

Using (x − a)² + (y − b)² = r²:

a = 3, b = −2 and r = 4.

(x − 3)² + (y − (−2))² = 4²

(x − 3)² + (y + 2)² = 16

Equation: (x − 3)² + (y + 2)² = 16.

3. Identifying the Centre and Radius

When a circle is given in standard form, its centre and radius can be identified directly.

Consider:

(x − 4)² + (y + 3)² = 49

Comparing with:

(x − a)² + (y − b)² = r²

• Centre = (4, −3)
• Radius = √49 = 7 units

Important sign rule: The signs inside the brackets are opposite to the signs of the centre coordinates.

• (x − 5)² gives x-coordinate 5.
• (x + 5)² gives x-coordinate −5.
• (y − 2)² gives y-coordinate 2.
• (y + 2)² gives y-coordinate −2.

Worked Example 3

Find the centre and radius of the circle:

(x + 2)² + (y − 5)² = 36

Solution

Comparing with the standard equation:

a = −2 and b = 5.

r² = 36, so r = 6.

Centre = (−2, 5); radius = 6 units.

Worked Example 4

A circle has centre (−4, −3) and passes through the point P(2, 5). Find its equation.

Solution

First find the radius using the distance formula:

r = √[(2 − (−4))² + (5 − (−3))²]

r = √[6² + 8²]

r = √100 = 10.

Using the standard equation:

(x + 4)² + (y + 3)² = 10²

Equation: (x + 4)² + (y + 3)² = 100.

4. The General Equation of a Circle

The standard equation can be expanded to give another useful form.

Start with:

(x − a)² + (y − b)² = r²

Expanding:

x² − 2ax + a² + y² − 2by + b² = r²

Collecting all terms on the left:

GENERAL EQUATION OF A CIRCLE

x² + y² + 2gx + 2fy + c = 0

The coefficients of x² and y² are equal, and there is no xy term.

Comparing the expanded equation with the general form:

• g = −a
• f = −b
• c = a² + b² − r²

Therefore, when the equation is in this form:

Centre = (−g, −f)

Radius = √(g² + f² − c)

These results will be established by completing the square in the next section.

5. Completing the Square

Completing the square is an algebraic method used to rewrite a quadratic expression in the form of a perfect square plus or minus a constant.

For example:

x² + 6x = (x + 3)² − 9

To complete the square:

1. Take half the coefficient of x.
2. Square the result.
3. Add and subtract that square.
4. Write the first three terms as a perfect square.

In general:

x² + px = (x + p/2)² − (p/2)²

5.1 Completing the Square in a Circle Equation

When a circle is given in expanded form, complete the square separately for the x-terms and the y-terms.

Worked Example 5

Find the centre and radius of the circle:

x² + y² − 6x + 4y − 12 = 0

Solution

Group the x-terms and y-terms:

(x² − 6x) + (y² + 4y) = 12

Complete the square for the x-terms:

x² − 6x = (x − 3)² − 9

Complete the square for the y-terms:

y² + 4y = (y + 2)² − 4

Substitute:

(x − 3)² − 9 + (y + 2)² − 4 = 12

Collect the constants:

(x − 3)² + (y + 2)² = 12 + 9 + 4

(x − 3)² + (y + 2)² = 25

Comparing with the standard equation:

Centre = (3, −2); radius = 5 units.

Worked Example 6

Find the centre and radius of:

x² + y² + 8x − 10y + 16 = 0

Solution

Group the terms:

(x² + 8x) + (y² − 10y) = −16

Complete the square:

x² + 8x = (x + 4)² − 16

y² − 10y = (y − 5)² − 25

Substitute:

(x + 4)² − 16 + (y − 5)² − 25 = −16

Rearrange:

(x + 4)² + (y − 5)² = 25

Centre = (−4, 5); radius = 5 units.

Worked Example 7: When the Coefficients Are Not 1

Find the centre and radius of:

2x² + 2y² + 8x − 12y + 8 = 0

Solution

First divide every term by 2:

x² + y² + 4x − 6y + 4 = 0

Group the terms:

(x² + 4x) + (y² − 6y) = −4

Complete the square:

x² + 4x = (x + 2)² − 4

y² − 6y = (y − 3)² − 9

Substitute:

(x + 2)² − 4 + (y − 3)² − 9 = −4

Therefore:

(x + 2)² + (y − 3)² = 9

Centre = (−2, 3); radius = 3 units.

Important Examination Note

If the coefficients of x² and y² are equal but not 1, divide the entire equation by that common coefficient before completing the square.

If the coefficients of x² and y² are unequal, or an xy term is present, the equation is not in the ordinary circle form covered by this method.

6. Conditions for an Equation to Represent a Real Circle

After completing the square, a circle equation should have the form:

(x − a)² + (y − b)² = r²

For a real circle with a positive radius:

r² must be greater than 0.

• If r² > 0, the equation represents a real circle.
• If r² = 0, it represents a single point (a circle of zero radius).
• If r² < 0, there are no real points satisfying the equation.

Worked Example 8

Determine the values of k for which the equation

x² + y² − 4x + 6y + k = 0

represents a real circle with positive radius.

Solution

Group the terms:

(x² − 4x) + (y² + 6y) = −k

Complete the square:

(x − 2)² − 4 + (y + 3)² − 9 = −k

Therefore:

(x − 2)² + (y + 3)² = 13 − k

For a real circle with positive radius:

13 − k > 0

k < 13.

7. Equation of a Circle Given the Endpoints of a Diameter

If the endpoints of a diameter are known, the centre is the midpoint of those endpoints.

The radius is half the length of the diameter.

Suppose the endpoints are A(x₁, y₁) and B(x₂, y₂).

1. Find the midpoint of AB to obtain the centre.
2. Find the length AB using the distance formula.
3. Divide AB by 2 to obtain the radius.
4. Substitute the centre and radius into the standard circle equation.

Worked Example 9

Find the equation of the circle whose diameter has endpoints A(−2, 1) and B(4, 5).

Solution

Step 1: Find the centre.

C = ((−2 + 4)/2, (1 + 5)/2)

C = (1, 3)

Step 2: Find the diameter.

AB = √[(4 + 2)² + (5 − 1)²]

AB = √(36 + 16)

AB = √52 = 2√13

Step 3: Find the radius.

r = √13

Step 4: Write the equation.

(x − 1)² + (y − 3)² = (√13)²

Equation: (x − 1)² + (y − 3)² = 13.

8. Equation of a Circle Passing Through Given Points

Sometimes a question gives points on a circle but does not give its centre or radius.

We can use the general equation:

x² + y² + 2gx + 2fy + c = 0

Substitute the coordinates of each given point into the equation. This produces simultaneous equations in g, f and c.

Three suitable points can be used to determine the three unknown constants, provided the points are not collinear.

Worked Example 10

Find the equation of the circle passing through A(0, 0), B(4, 0) and C(0, 6).

Solution

Start with the general equation:

x² + y² + 2gx + 2fy + c = 0

Using A(0, 0):

0 + 0 + 0 + 0 + c = 0

Therefore c = 0.

Using B(4, 0):

4² + 0 + 8g + 0 + 0 = 0

16 + 8g = 0

g = −2.

Using C(0, 6):

0 + 6² + 0 + 12f + 0 = 0

36 + 12f = 0

f = −3.

Substitute g = −2, f = −3 and c = 0:

x² + y² − 4x − 6y = 0.

Completing the square gives:

(x − 2)² + (y − 3)² = 13.

Therefore:

Centre = (2, 3); radius = √13 units.

9. Determining the Position of a Point Relative to a Circle

We can determine whether a point lies inside, on, or outside a circle by comparing its distance from the centre with the radius.

For a circle with centre C(a, b) and radius r, calculate:

d² = (x − a)² + (y − b)²

• If d² < r², the point lies inside the circle.
• If d² = r², the point lies on the circumference.
• If d² > r², the point lies outside the circle.

Worked Example 11

A circle has equation:

(x − 2)² + (y − 1)² = 25

Determine the position of P(5, 5) relative to the circle.

Solution

Centre = (2, 1) and r² = 25.

For P(5, 5):

d² = (5 − 2)² + (5 − 1)²

d² = 9 + 16 = 25.

Since d² = r²:

P lies on the circumference.

Worked Example 12

Using the same circle, determine whether Q(2, 3) is inside or outside it.

Solution

d² = (2 − 2)² + (3 − 1)²

d² = 0 + 4 = 4.

Since 4 < 25:

Q lies inside the circle.

10. Common Examination Errors

1. Incorrect signs: Remember that the centre coordinates have signs opposite to those inside the brackets.
2. Forgetting to square the radius: The right-hand side of the standard equation is r², not r.
3. Incomplete square completion: Complete the square for both x and y terms.
4. Forgetting to divide: If x² and y² have a common coefficient other than 1, divide the entire equation first.
5. Incorrect midpoint: The centre is the midpoint of the diameter endpoints, not one of the endpoints.
6. Incorrect radius: When given a diameter, divide its length by 2.
7. Not checking the radius: A real circle with positive radius requires r² > 0.

11. Summary of Important Formulae

Standard equation: (x − a)² + (y − b)² = r²

Centre at origin: x² + y² = r²

General equation: x² + y² + 2gx + 2fy + c = 0

Centre from general form: (−g, −f)

Radius from general form: √(g² + f² − c)

Diameter: 2r

Radius: Half the diameter

Midpoint of diameter: ((x₁ + x₂)/2, (y₁ + y₂)/2)

Position of point: Compare (x − a)² + (y − b)² with r².

12. Examination Practice

Attempt the following questions independently. Show all working and leave your answers in exact form where appropriate.

Exercise 1: Basic Circle Equations

1. Find the equation of a circle with centre (2, −3) and radius 5.
2. Find the centre and radius of (x + 4)² + (y − 2)² = 49.
3. Find the equation of a circle centred at the origin and passing through (6, 8).
4. A circle has centre (−1, 3) and passes through (2, 7). Find its equation.
5. Find the centre and radius of x² + y² − 8x + 6y + 9 = 0.

Exercise 2: Completing the Square

6. Find the centre and radius of x² + y² + 4x − 10y + 13 = 0.
7. Find the centre and radius of x² + y² − 2x − 8y − 8 = 0.
8. Find the centre and radius of 3x² + 3y² − 12x + 18y − 12 = 0.
9. Determine the values of k for which x² + y² + 6x − 4y + k = 0 represents a real circle with positive radius.
10. Express (x − 5)² + (y + 2)² = 36 in expanded general form.

Exercise 3: Finding Circle Equations

11. Find the equation of a circle whose diameter has endpoints A(0, 0) and B(6, 8).
12. Find the equation of a circle whose diameter has endpoints A(−4, 2) and B(2, 10).
13. Find the equation of the circle passing through (0, 0), (6, 0) and (0, 8).
14. A circle has centre (3, −1) and passes through (7, 2). Find its equation in expanded form.

Exercise 4: Further Applications

15. Determine whether the point (4, 5) lies inside, on, or outside the circle (x − 1)² + (y − 1)² = 25.
16. Determine whether the point (−2, 3) lies inside, on, or outside the circle x² + y² = 13.
17. Find the radius of the circle x² + y² + 4x − 6y + k = 0 in terms of k. State the condition for a real circle with positive radius.
18. A circle has centre (a, 2) and passes through (4, 6). If its radius is 5, find the possible values of a.

13. Worked Exercise Solutions

Try each question first. Open the relevant solution section to check your work.

Solutions 1–5

1. Centre (2, −3), radius 5. Equation: (x − 2)² + (y + 3)² = 25.

2. Centre = (−4, 2); radius = √49 = 7 units.

3. r² = 6² + 8² = 100. Equation: x² + y² = 100.

4. r² = (2 + 1)² + (7 − 3)² = 9 + 16 = 25. Equation: (x + 1)² + (y − 3)² = 25.

5. Completing the square gives (x − 4)² + (y + 3)² = 16. Centre = (4, −3); radius = 4 units.

Solutions 6–10

6. (x + 2)² + (y − 5)² = 16. Centre = (−2, 5); radius = 4 units.

7. (x − 1)² + (y − 4)² = 25. Centre = (1, 4); radius = 5 units.

8. Divide by 3 first: x² + y² − 4x + 6y − 4 = 0. Completing the square gives (x − 2)² + (y + 3)² = 17. Centre = (2, −3); radius = √17 units.

9. Completing the square gives (x + 3)² + (y − 2)² = 13 − k. For positive radius, k < 13.

10. Expanding gives x² − 10x + 25 + y² + 4y + 4 = 36. Therefore x² + y² − 10x + 4y − 7 = 0.

Solutions 11–14

11. Centre = (3, 4), radius = 5. Equation: (x − 3)² + (y − 4)² = 25.

12. Centre = (−1, 6). Diameter² = 6² + 8² = 100, so radius = 5. Equation: (x + 1)² + (y − 6)² = 25.

13. General equation: x² + y² + 2gx + 2fy + c = 0. The origin gives c = 0. The point (6, 0) gives g = −3, while (0, 8) gives f = −4. Hence x² + y² − 6x − 8y = 0.

14. r² = (7 − 3)² + (2 + 1)² = 25. Standard form: (x − 3)² + (y + 1)² = 25. Expanding: x² + y² − 6x + 2y − 15 = 0.

Solutions 15–18

15. d² = (4 − 1)² + (5 − 1)² = 9 + 16 = 25. Since d² = r², the point is on the circumference.

16. d² = (−2)² + 3² = 13. Since d² = r², the point is on the circumference.

17. Completing the square gives (x + 2)² + (y − 3)² = 13 − k. Thus r = √(13 − k), and a positive radius requires k < 13.

18. r² = (4 − a)² + (6 − 2)² = 25. Therefore (4 − a)² + 16 = 25, so (4 − a)² = 9. Hence a = 1 or a = 7.

End of Part 2

You should now be able to form circle equations, identify centres and radii, complete the square, and solve problems involving points and circles.

Practise the algebra carefully. Completing the square is especially important because it appears in many coordinate geometry problems.

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