Chapter 05: Circles – Part 4: Chords, Intersecting Circles and Examination Practice

MATHEMATICS HIPHOP ACADEMY

CHAPTER 05: CIRCLES

PART 4: CHORDS, INTERSECTING CIRCLES AND EXAMINATION PRACTICE

A-Level Pure Mathematics

1. Introduction

A chord is a straight line segment joining two points on the circumference of a circle. Chords have important geometric properties that allow us to solve problems involving lengths, angles, distances and intersecting circles.

In this final part, we shall investigate the relationship between chords and the centre of a circle, learn how to calculate chord lengths, study intersecting chords and examine the common chord of two circles.

Learning objectives

• Understand the properties of chords.
• Find the perpendicular distance from the centre to a chord.
• Calculate the length of a chord.
• Apply the intersecting chords theorem.
• Find the common chord of two intersecting circles.
• Solve examination questions involving circles and chords.

2. Important Properties of Chords

A chord is a line segment whose endpoints lie on the circumference of a circle.

The diameter is the longest chord of a circle because it passes through the centre.

Several important results follow from the geometry of chords.

Property 1: Perpendicular from the centre to a chord

The perpendicular drawn from the centre of a circle to a chord bisects the chord.

If OM is perpendicular to chord AB, then:

AM = MB

Property 2: Equal chords

Equal chords of the same circle are equidistant from the centre.

Conversely, chords that are equidistant from the centre of the same circle are equal.

Property 3: The perpendicular bisector of a chord

The perpendicular bisector of a chord passes through the centre of the circle.

This property is useful when the centre of a circle is unknown but two chords are given.

Property 4: The angle at the centre

The angle subtended by a chord at the centre is twice the angle subtended by the same chord at any point on the remaining circumference.

This is useful in geometrical questions involving angles and chords.

3. Finding the Length of a Chord

Consider a circle with radius r. Let the perpendicular distance from its centre to a chord be d, and let the chord have length L.

The perpendicular from the centre bisects the chord. Therefore, half the chord has length L/2.

Using Pythagoras' theorem:

(L/2)² + d² = r²

Rearranging:

(L/2)² = r² − d²

L = 2√(r² − d²)

Length of a chord

Where:

• L is the chord length.
• r is the radius of the circle.
• d is the perpendicular distance from the centre to the chord.

Worked Example 1

A circle has radius 13 cm. A chord is 5 cm from the centre. Find the length of the chord.

Solution

Given:

r = 13 cm and d = 5 cm.

Using:

L = 2√(r² − d²)

L = 2√(13² − 5²)

L = 2√(169 − 25)

L = 2√144

L = 2 × 12

L = 24 cm

Final answer

4. Chords in the Cartesian Plane

Coordinate geometry allows us to find chord lengths and distances from the centre using equations.

Consider the circle:

x² + y² = r²

Its centre is (0, 0) and its radius is r.

If a horizontal chord is represented by y = k, substitution gives:

x² + k² = r²

x² = r² − k²

x = ±√(r² − k²)

The two endpoints are therefore:

(−√(r² − k²), k)

(√(r² − k²), k)

The chord length is the difference between the two x-coordinates:

L = 2√(r² − k²)

Worked Example 2

The line y = 3 cuts the circle:

x² + y² = 25

at points P and Q. Find the coordinates of P and Q and the length of chord PQ.

Solution

Substitute y = 3 into the circle:

x² + 3² = 25

x² + 9 = 25

x² = 16

x = ±4.

When x = −4, the point is (−4, 3).

When x = 4, the point is (4, 3).

Therefore:

P = (−4, 3), Q = (4, 3).

Length of PQ:

PQ = 4 − (−4)

PQ = 8 units.

P(−4, 3), Q(4, 3)

PQ = 8 units

5. The Intersecting Chords Theorem

When two chords intersect inside a circle, the products of the lengths of the two segments of one chord and the two segments of the other chord are equal.

Suppose chords AB and CD intersect at P inside a circle.

PA × PB = PC × PD

Intersecting Chords Theorem

This result is also known as the chord-chord theorem.

Worked Example 3

Two chords AB and CD intersect at P inside a circle. Given that PA = 3 cm, PB = 8 cm and PC = 4 cm, find PD.

Solution

Using the intersecting chords theorem:

PA × PB = PC × PD

Substitute the values:

3 × 8 = 4 × PD

24 = 4PD

PD = 24/4

PD = 6 cm

6. Two Circles Intersecting

Two circles may be positioned in several ways:

• They may be completely separate.
• They may touch externally at one point.
• They may intersect at two points.
• One circle may lie entirely inside another without touching.
• They may touch internally at one point.
• They may have the same centre and radius, in which case they coincide.

When two distinct circles intersect at two points, the line segment joining the two intersection points is called their common chord.

7. The Common Chord of Two Circles

A particularly important result is that the line joining the centres of two intersecting circles is perpendicular to their common chord and bisects it.

This means that if the centres are O₁ and O₂, and the circles intersect at A and B, then O₁O₂ is perpendicular to AB and passes through the midpoint of AB.

Important result

The common chord is perpendicular to the line joining the centres of the two circles.

The line joining the centres bisects the common chord.

8. Finding the Equation of a Common Chord

Suppose two circles are given by:

S₁ = 0

S₂ = 0

At their common points, both equations are satisfied.

Subtracting the two circle equations eliminates the squared terms x² and y². The resulting equation is linear and represents the line through their common points.

Method

1. Write down the two circle equations.
2. Subtract one equation from the other.
3. Simplify to obtain the equation of the common chord.
4. Substitute the common chord equation into either circle.
5. Solve for the coordinates of the two intersection points.

Worked Example 4

Find the equation of the common chord of the circles:

x² + y² = 25

(x − 6)² + y² = 25

Also find their points of intersection.

Solution

Expand the second equation:

x² − 12x + 36 + y² = 25

x² + y² − 12x + 11 = 0.

The first equation is:

x² + y² − 25 = 0.

Subtract the first equation from the second:

−12x + 36 = 0

12x = 36

x = 3.

Therefore, the common chord has equation:

x = 3

To find the intersection points, substitute x = 3 into the first circle:

3² + y² = 25

9 + y² = 25

y² = 16

y = ±4.

Hence the two points of intersection are:

(3, 4) and (3, −4)

Common chord: x = 3

The length of the common chord is:

4 − (−4) = 8 units.

9. The Radical Axis

The common chord line obtained by subtracting the equations of two intersecting circles is called their radical axis.

The radical axis is the locus of points that have equal powers with respect to the two circles.

For two intersecting circles, the radical axis passes through both intersection points. It is therefore the equation of their common chord.

Even when two circles do not intersect at real points, their radical axis can still be found algebraically by subtracting their equations.

10. Examination Strategy

Circle questions often combine several ideas from earlier parts of this chapter. A question may require completing the square, finding a tangent, calculating a chord length or determining the common chord of two circles.

A reliable examination approach

1. Identify the centre and radius of each circle.
2. Write equations clearly before substituting.
3. Use the discriminant when determining the number of intersections or a tangent condition.
4. Use perpendicular gradients when finding a tangent at a known point.
5. Use Pythagoras' theorem for chord-length questions.
6. Subtract circle equations to find a common chord.
7. Show all algebraic steps and state the final answer with appropriate units.

11. Examination Practice

Attempt these questions independently. Show complete working and avoid looking at the answers until you have finished.

Question 1

A circle has equation:

x² + y² = 169.

A chord is 5 units from the centre. Find the length of the chord.

Question 2

The line y = 4 intersects the circle:

x² + y² = 100.

Find the coordinates of the intersection points and the length of the chord cut off by the line.

Question 3

Two chords AB and CD intersect at P inside a circle. If PA = 5 cm, PB = 12 cm and PC = 6 cm, find PD.

Question 4

Find the equation of the common chord of the circles:

x² + y² = 25

and

(x − 8)² + y² = 25.

Hence find their points of intersection.

Question 5

A circle has centre (2, −1) and radius 5. A horizontal chord is 3 units above the centre.

Find the equation of the chord and its length.

Question 6

Two circles have equations:

x² + y² − 4x − 6y + 9 = 0

and

x² + y² + 2x − 2y − 3 = 0.

Find the equation of their radical axis.

Question 7

A circle has centre O and radius 10 cm. A chord AB is 6 cm from O.

Find the length of AB and the area of triangle OAB.

12. Worked Answers to Examination Practice

Use the expandable solutions below to check your work.

Question 1 – Show solution

Radius r = 13 and perpendicular distance d = 5.

Chord length L = 2√(r² − d²)

L = 2√(169 − 25)

L = 2√144

Answer: 24 units.

Question 2 – Show solution

Substitute y = 4:

x² + 16 = 100

x² = 84

x = ±√84 = ±2√21.

The points are (−2√21, 4) and (2√21, 4).

Chord length = 2√21 − (−2√21)

Answer: 4√21 units.

Question 3 – Show solution

Using PA × PB = PC × PD:

5 × 12 = 6 × PD

60 = 6PD

Answer: PD = 10 cm.

Question 4 – Show solution

First circle:

x² + y² − 25 = 0.

Expand the second circle:

(x − 8)² + y² = 25

x² + y² − 16x + 39 = 0.

Subtract the first equation from the second:

−16x + 64 = 0

Common chord: x = 4.

Substitute x = 4 into the first circle:

16 + y² = 25

y² = 9

y = ±3.

Points of intersection: (4, 3) and (4, −3).

Question 5 – Show solution

The centre is (2, −1).

Three units above the centre means:

y = −1 + 3 = 2.

Therefore the chord equation is y = 2.

The perpendicular distance from the centre to the chord is 3 units.

Using the chord formula:

L = 2√(25 − 9)

L = 2√16

Answer: Chord equation y = 2; length = 8 units.

Question 6 – Show solution

Subtract the second equation from the first:

(−4x − 6y + 9) − (2x − 2y − 3) = 0

−6x − 4y + 12 = 0

Divide by −2:

Answer: 3x + 2y − 6 = 0.

Question 7 – Show solution

Radius r = 10 cm and perpendicular distance d = 6 cm.

Half the chord is:

√(10² − 6²) = √64 = 8 cm.

Therefore AB = 2 × 8 = 16 cm.

Area of triangle OAB:

= ½ × base × perpendicular height

= ½ × 16 × 6

Answer: AB = 16 cm; area = 48 cm².

13. Final Chapter 05 Summary

• The standard equation of a circle is (x − a)² + (y − b)² = r².
• The general equation can be converted to standard form by completing the square.
• A line may intersect a circle at two points, one point or no real points.
• The discriminant determines the nature of the intersection.
• A tangent is perpendicular to the radius at the point of contact.
• A perpendicular from the centre to a chord bisects the chord.
• The length of a chord is L = 2√(r² − d²).
• Intersecting chords satisfy PA × PB = PC × PD.
• The line joining the centres of two intersecting circles bisects their common chord at right angles.
• Subtracting two circle equations gives the radical axis.

CHAPTER 05 COMPLETED!

CIRCLES

You have completed all four parts of the Circles chapter.

Revise the worked examples, attempt the exercises again and make sure you understand the methods before moving to the next chapter.

MATHEMATICS HIPHOP ACADEMY

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A-Level Pure Mathematics | Chapter 05 – Circles