Chapter 05: Circles – Part 3: Lines and Circles – Intersections and Tangents

MATHEMATICS HIPHOP ACADEMY

CHAPTER 05: CIRCLES

PART 3: LINES AND CIRCLES – INTERSECTIONS AND TANGENTS

A-Level Pure Mathematics

1. Introduction

A straight line may meet a circle at two points, touch it at exactly one point, or fail to meet it altogether. These possibilities can be investigated using simultaneous equations and the discriminant of a quadratic equation.

We shall also study the important relationship between a tangent and the radius drawn to its point of contact.

Learning objectives

• Find the points where a straight line meets a circle.
• Use the discriminant to determine the nature of an intersection.
• Find conditions for a line to be a tangent.
• Find the equation of a tangent at a given point.
• Solve examination questions involving lines and circles.

2. How a Straight Line Meets a Circle

There are three possible relationships between a straight line and a circle.

Case 1: Two points of intersection

A line passing through a circle at two distinct points is called a secant. The portion of the line inside the circle is a chord.

There are two distinct solutions when the resulting quadratic equation has a positive discriminant.

Case 2: One point of intersection

A line touching a circle at exactly one point is called a tangent.

The tangent has one repeated point of intersection with the circle. The resulting quadratic equation has a discriminant equal to zero.

Case 3: No point of intersection

A line outside the circle does not meet the circumference.

The resulting quadratic equation has a negative discriminant and therefore has no real solutions.

3. The Discriminant Test

Suppose the equations of a line and a circle are given. We substitute the line equation into the circle equation to obtain a quadratic equation:

ax² + bx + c = 0

Δ = b² − 4ac

The discriminant determines the number of real points of intersection.

Discriminant	Number of points	Relationship
Δ > 0	Two	Secant
Δ = 0	One	Tangent
Δ < 0	None	No intersection

Important examination principle

When a question asks you to prove that a line is tangent to a circle, show that substitution produces a quadratic equation whose discriminant is zero.

4. Finding Points of Intersection

To find the points where a straight line meets a circle:

1. Write down the equation of the circle and the equation of the line.
2. Make one variable the subject of the line equation.
3. Substitute into the circle equation.
4. Expand and simplify to obtain a quadratic equation.
5. Solve the quadratic equation.
6. Substitute each value back into the line equation to find the corresponding second coordinate.
7. State both points of intersection.

Worked Example 1

Find the points of intersection of the line:

y = x + 1

and the circle:

x² + y² = 25

Solution

Substitute y = x + 1 into the circle equation:

x² + (x + 1)² = 25

x² + x² + 2x + 1 = 25

2x² + 2x − 24 = 0

Divide throughout by 2:

x² + x − 12 = 0

Factorise:

(x + 4)(x − 3) = 0

Therefore:

x = −4 or x = 3.

When x = −4:

y = −4 + 1 = −3.

When x = 3:

y = 3 + 1 = 4.

Final answer

The points of intersection are:

(−4, −3) and (3, 4)

5. Finding the Condition for a Tangent

When a line is tangent to a circle, it touches the circle at exactly one point. Therefore, after substitution, the quadratic equation has equal roots.

For a quadratic equation:

ax² + bx + c = 0

the condition for equal roots is:

b² − 4ac = 0

This is the tangent condition.

Worked Example 2

Find the values of k for which the line:

y = x + k

is a tangent to the circle:

x² + y² = 8.

Solution

Substitute y = x + k into the circle:

x² + (x + k)² = 8

x² + x² + 2kx + k² = 8

2x² + 2kx + k² − 8 = 0

Comparing with ax² + bx + c = 0:

• a = 2
• b = 2k
• c = k² − 8

For tangency, the discriminant must equal zero:

b² − 4ac = 0

(2k)² − 4(2)(k² − 8) = 0

4k² − 8k² + 64 = 0

−4k² + 64 = 0

k² = 16

Therefore:

k = 4 or k = −4

Final answer

The two possible tangent lines are:

y = x + 4

y = x − 4

6. The Relationship Between a Tangent and a Radius

A very important property of circles is that a tangent is perpendicular to the radius drawn to the point of contact.

Tangent ⟂ Radius

The angle between the tangent and the radius at the point of contact is 90°.

This property allows us to find the equation of a tangent using gradients.

If the gradient of the radius is m₁ and the gradient of the tangent is m₂, then:

m₁ × m₂ = −1

Therefore, where the radius has a non-zero finite gradient:

Gradient of tangent = −1 ÷ gradient of radius

Remember that a horizontal radius has a vertical tangent, while a vertical radius has a horizontal tangent.

7. Equation of a Tangent at a Given Point

To find the equation of a tangent at a given point on a circle:

1. Find the centre of the circle.
2. Identify the point of contact.
3. Calculate the gradient of the radius joining the centre to that point.
4. Find the negative reciprocal to obtain the tangent gradient, where applicable.
5. Use the straight-line equation y − y₁ = m(x − x₁).
6. Simplify your answer.

Worked Example 3

Find the equation of the tangent to the circle:

x² + y² = 25

at the point (3, 4).

Solution

The centre of the circle is C(0, 0).

The point of contact is P(3, 4).

Gradient of the radius CP:

m₁ = (4 − 0)/(3 − 0)

m₁ = 4/3

Since the tangent is perpendicular to the radius:

m₂ = −3/4

Using the point-gradient equation:

y − y₁ = m(x − x₁)

y − 4 = −3/4(x − 3)

Multiply throughout by 4:

4y − 16 = −3x + 9

Rearranging:

3x + 4y = 25

Equation of the tangent

8. Finding the Point of Contact

Sometimes a question gives a tangent line and asks for the point where it touches the circle.

One method is to substitute the tangent equation into the circle equation. Since the line is tangent, the resulting quadratic has a repeated root. Find that root and substitute it into the line equation to obtain the point of contact.

Worked Example 4

The line y = x + 4 is tangent to the circle x² + y² = 8. Find the point of contact.

Solution

Substitute y = x + 4:

x² + (x + 4)² = 8

2x² + 8x + 8 = 0

Divide by 2:

x² + 4x + 4 = 0

(x + 2)² = 0

Therefore x = −2.

Substitute into y = x + 4:

y = −2 + 4 = 2.

Point of contact = (−2, 2)

9. Using the Perpendicular Distance from a Centre to a Line

There is another useful way of checking whether a line is tangent to a circle.

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

Ax + By + C = 0

The perpendicular distance d from the centre to the line is:

d = |Aa + Bb + C| / √(A² + B²)

Compare this distance with the radius:

• If d < r, the line cuts the circle at two points.
• If d = r, the line is a tangent.
• If d > r, the line does not meet the circle.

Worked Example 5

Determine whether the line 3x + 4y − 25 = 0 is tangent to the circle x² + y² = 25.

Solution

The centre is (0, 0), and the radius is 5.

The perpendicular distance is:

d = |3(0) + 4(0) − 25| / √(3² + 4²)

d = 25/√25

d = 25/5 = 5.

Since d = r = 5, the line is a tangent to the circle.

Conclusion: The line is tangent to the circle.

10. Common Examination Mistakes

Mistake 1: Forgetting to square the entire expression after substitution.

Correct: (x + 3)² = x² + 6x + 9.

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Mistake 2: Using a positive reciprocal instead of a negative reciprocal when finding the gradient of a perpendicular line.

Remember: perpendicular gradients have product −1, provided both gradients are finite.

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Mistake 3: Saying a tangent has two distinct points of intersection.

A tangent has exactly one point of contact, represented by a repeated root.

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Mistake 4: Giving only the x-coordinate when asked for points of intersection.

Always substitute each x-value back into the line equation to find its corresponding y-coordinate.

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Mistake 5: Using the gradient of the radius as the gradient of the tangent.

The tangent is perpendicular to the radius, not parallel to it.

11. Practice Exercises

Attempt all the questions before opening the solutions. Show your working clearly, especially when using the discriminant.

Exercise 1

Find the points of intersection of:

y = 2x − 3

and (x − 3)² + (y + 2)² = 25.

Exercise 2

Find the points of intersection of the line y = x + 2 and the circle x² + y² = 20.

Exercise 3

Find the values of k for which the line y = x + k is a tangent to the circle x² + y² = 8.

Exercise 4

Find the equation of the tangent to the circle x² + y² = 25 at the point (3, 4).

Exercise 5

A circle has centre (2, −1) and radius 5. Find the points where the vertical line x = 5 meets the circle.

Exercise 6

The line y = 3 cuts the circle x² + y² = 25 at P and Q. Find the coordinates of P and Q and calculate the length of chord PQ.

12. Worked Answers to Practice Exercises

Open each solution only after attempting the corresponding question.

Exercise 1 – Show solution

Substitute y = 2x − 3:

(x − 3)² + (2x − 1)² = 25

x² − 6x + 9 + 4x² − 4x + 1 = 25

5x² − 10x − 15 = 0

x² − 2x − 3 = 0

(x − 3)(x + 1) = 0

x = 3 or x = −1.

When x = 3, y = 3.

When x = −1, y = −5.

Answer: (3, 3) and (−1, −5).

Exercise 2 – Show solution

x² + (x + 2)² = 20

2x² + 4x − 16 = 0

x² + 2x − 8 = 0

(x + 4)(x − 2) = 0

x = −4 or x = 2.

When x = −4, y = −2.

When x = 2, y = 4.

Answer: (−4, −2) and (2, 4).

Exercise 3 – Show solution

x² + (x + k)² = 8

2x² + 2kx + k² − 8 = 0.

For tangency, b² − 4ac = 0.

(2k)² − 4(2)(k² − 8) = 0

4k² − 8k² + 64 = 0

k² = 16

Answer: k = 4 or k = −4.

Exercise 4 – Show solution

The centre is (0, 0).

Gradient of radius = 4/3.

Gradient of tangent = −3/4.

Using y − 4 = −3/4(x − 3):

4y − 16 = −3x + 9

Answer: 3x + 4y = 25.

Exercise 5 – Show solution

The circle is:

(x − 2)² + (y + 1)² = 25.

Substitute x = 5:

(5 − 2)² + (y + 1)² = 25

9 + (y + 1)² = 25

(y + 1)² = 16

y + 1 = ±4

y = 3 or y = −5.

Answer: (5, 3) and (5, −5).

Exercise 6 – Show solution

Substitute y = 3 into x² + y² = 25:

x² + 9 = 25

x² = 16

x = ±4.

Therefore P(−4, 3) and Q(4, 3).

Since the chord is horizontal, its length is:

PQ = 4 − (−4) = 8.

Answer: P(−4, 3), Q(4, 3), and PQ = 8 units.

13. Chapter Summary

• A secant meets a circle at two distinct points.
• A tangent touches a circle at exactly one point.
• A line that misses a circle has no real points of intersection.
• Substitution of a line into a circle produces a quadratic equation.
• Positive discriminant means two intersections.
• Zero discriminant means tangency.
• Negative discriminant means no real intersection.
• A tangent is perpendicular to the radius at the point of contact.
• The perpendicular distance from the centre to a tangent equals the radius.

MATHEMATICS HIPHOP ACADEMY

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Chapter 05 – Circles | Part 3 Completed

Proceed to Part 4: Chords, Intersecting Circles and Examination Practice.