Circles and tangents Quiz (Grade 8+)

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Circles and tangents — Notes & Worked Examples Show / Hide

Explanation:

At GCSE, a circle with centre (0, 0) and radius r has equation x² + y² = r². Because a tangent to a circle is always perpendicular to the radius at the point of contact P(x₁, y₁), their gradients multiply to give −1.

Follow these 3 steps to find the equation of the tangent at P(x₁, y₁): (1) find the gradient of the radius OP using m_r = y₁x₁; (2) find the perpendicular tangent gradient m_t = −1m_r; (3) substitute m_t and (x₁, y₁) into y − y₁ = m_t(x − x₁).

Example 1: Find the equation of the tangent to x² + y² = 25 at the point P(3, −4).

Solution:

Gradient of radius OP: m_r = −4 − 03 − 0 = −43

Gradient of tangent (perpendicular): m_t = +34

Equation: y − (−4) = 34(x − 3) ⇒ 4y + 16 = 3x − 9

Answer: y = 34x − 254 (or 3x − 4y = 25)

Example 2: The tangent to x² + y² = 20 at A(2, 4) crosses the x-axis at B. Find the coordinates of B.

Solution:

Gradient of radius OA: m_r = 42 = 2 ⇒ m_t = −12

Tangent: y − 4 = −12(x − 2) ⇒ y = −12x + 5

On the x-axis, y = 0: −12x + 5 = 0 ⇒ x = 10

Answer: B(10, 0)

One Page Guide (PDF) Circles and tangents Worksheets

Question 1

[1 mark]

The point P(3, 4) lies on the circle x² + y² = 25.
Find the equation of the tangent to the circle at P in the form ax + by = c.

Worked Solution & Answer

Gradient of radius OP = 43, so tangent gradient = -34 ⇒ 3x + 4y = 25. B1

Answer: 3x + 4y = 25

Question 2

[1 mark]

The point P(5, 12) lies on the circle x² + y² = 169.
Find the equation of the tangent to the circle at P in the form ax + by = c.

Worked Solution & Answer

Gradient of radius OP = 125, so tangent gradient = -512 ⇒ 5x + 12y = 169. B1

Answer: 5x + 12y = 169

Question 3 Specimen Set 1 2H Q23

[3 marks]

Here is a circle, centre O, and the tangent to the circle at the point P(4, 3) on the circle.

Find an equation of the tangent at the point P.

y =

Worked Solution & Answer

Gradient of radius OP = 3/4. Gradient of tangent = –4/3. Equation: y – 3 = –(4/3)(x – 4) M2
⇒ y = –(4/3)x + 25/3 (or 4x + 3y = 25). A1

Answer: y = y = -43x + 253

Question Paper (PDF) • Mark Scheme (PDF)

Question 4 AQA Jun 2017 1H Q27

[4 marks]

P(-1, 4) is a point on a circle, centre O

Work out the equation of the tangent to the circle at P.
Give your answer in the form y = mx + c

Answer

Worked Solution & Answer

Gradient of radius OP from (0, 0) to (-1, 4): M1
grad_OP = 4 - 0-1 - 0 = -4
The tangent is perpendicular to the radius at P, so: M1
m = -1-4 = 14
Equation through (-1, 4): M1
y - 4 = 14(x - (-1)) = 14x + 14
y = 14x + 174 (or y = 0.25x + 4.25) A1

Answer: Answer y = 14x + 174

Question Paper (PDF) • Mark Scheme (PDF)

Question 5 OCR Nov 2023 6H Q18

[4 marks]

A, B and C are points on the circumference of a circle.

Points C and D lie on a tangent to the circle at C.
Angle BCD = 45°.
Angle ACB = 58°.
AB = 8 cm.
Find the length of BC.

Worked Solution & Answer

(The circumference of a circle is π × diameter = 2πr)
By the alternate segment theorem, ∠ BAC = ∠ BCD = 45° M3
By the sine rule: BC\sin 45° = 8\sin 58° ⇒ BC = 8\sin 45°\sin 58° = 6.670… ≈ 6.67 cm. A1

Answer: 6.67 cm

Question Paper (PDF) • Mark Scheme (PDF)

Question 6 OCR Jun 2018 5H Q16

[4 marks]

A, B, C and D are points on the circumference of a circle.

PQ is a tangent to the circle at D.
Angle BDQ = 72° and angle ABD = 63°.

(a) Work out angle x.
Give a reason for your answer. [2 marks]

(b) Work out angle y.
Give a reason for your answer. [2 marks]

Worked Solution & Answer

(a)

(The circumference of a circle is π × diameter = 2πr)
By the alternate segment theorem, the angle between the tangent PQ and chord BD equals the angle subtended by BD in the alternate segment, so x = 72°. M1 C1

Answer (a): Angle x = 72° because angles in the alternate segment are equal.

(b)

(The circumference of a circle is π × diameter = 2πr)
ABCD is a cyclic quadrilateral, and opposite angles of a cyclic quadrilateral add up to 180°: M1
y = 180° - 72° = 108°. C1

Answer (b): Angle y = 108° because opposite angles of a cyclic quadrilateral add up to 180°.

Question Paper (PDF) • Mark Scheme (PDF)

Question 7 IGCSE Jun 2019 1H Q20

[5 marks]

A, B, C and D are points on a circle.

Work out the size of angle BCD.
Give a reason for each stage of your working.

TDV is the tangent to the circle at D.
AB = AD
Angle ADT = 71°

Worked Solution & Answer

Angle ABD = 71° (alternate segment theorem) M2
Angle ADB = 71° (base angles of an isosceles triangle are equal) M1
Angle BAD = 180° – 71° – 71° = 38° (angles in a triangle sum to 180°) M1
Angle BCD = 180° – 38° = 142° (opposite angles of a cyclic quadrilateral sum to 180°). A1

Answer: 142° (with full circle theorem reasons) °

Question Paper (PDF) • Mark Scheme (PDF)

Question 8

[5 marks]

(a) Draw graph of x² + y² = 12.25 (2) (b) Hence find estimates for solutions of x² + y² = 12.25 and 2x + y = 1 (3)

(a) On the grid, draw the graph of x² + y² = 12.25 [2 marks]

(b) Hence find estimates for the solutions of the simultaneous equations
x² + y² = 12.25
2x + y = 1 [3 marks]

Worked Solution & Answer

(a)

r = √12.25 = 3.5 M1 A1

Answer (a): Circle drawn with centre (0, 0) and radius 3.5

(b)

Draw line y = 1 – 2x M2
intersections with circle are at (2.0, –2.9) and (–1.2, 3.3) A1

Answer (b): x = 2.0, y = -2.9 and x = -1.2, y = 3.3

Question Paper (PDF) • Mark Scheme (PDF)

Question 9 OCR Nov 2020 6H Q16

[5 marks]

The diagram shows a circle, centre O.
Points A, B and C lie on the circumference of the circle.
Line AOB is a diameter.
Line DAE is a tangent to the circle.
Angle CAE = 32°.

(a) Give a reason why angle ACB is a right angle. [1 mark]

(b) The radius of the circle is 8 cm.
Calculate length BC. [4 marks]

Worked Solution & Answer

(a)

(The circumference of a circle is π × diameter = 2πr)
(The circumference of a circle is π × diameter = 2πr) C1

Answer (a): The angle in a semi-circle is 90°.

(b)

(The circumference of a circle is π × diameter = 2πr)
AB = 2 × 8 = 16 cm M2
Angle OAE = 90° (tangent perpendicular to radius), so angle BAC = 90° - 32° = 58°. M1
BC = 16 \sin 58° = 13.5687… ≈ 13.6 cm A1

Answer (b): 13.6 cm

Question Paper (PDF) • Mark Scheme (PDF)

Question 10 OCR Nov 2022 6H Q18

[9 marks]

A circle has equation x² + y² = 100.
The sketch shows the circle and two points, A and B, which lie on the circumference of the circle.

(a) Write down the coordinates of point A. [1 mark]

(b) Point B has x-coordinate 3.
Find the exact value of the y-coordinate of point B. [3 marks]

(c) Another point, C, lies on the circle and has a y-coordinate that is seven times its x-coordinate.
Find the two possible pairs of coordinates for point C.
Give your answers in exact form.
You must show your working. [5 marks]

Worked Solution & Answer

(a)

(The circumference of a circle is π × diameter = 2πr)
Radius r = √100 = 10, and A lies on the positive x-axis ⇒ (10, 0). B1

Answer (a): (10, 0)

(b)

(The circumference of a circle is π × diameter = 2πr)
3² + y² = 100 ⇒ y² = 91 M2
Since B is below the x-axis, y = −√91. A1

Answer (b): -√(91)

(c)

(The circumference of a circle is π × diameter = 2πr)
Substitute y = 7x into x² + y² = 100: M2
x² + (7x)² = 100 ⇒ x² + 49x² = 100 ⇒ 50x² = 100 ⇒ x² = 2 ⇒ x = ±√2. M1
When x = √2, y = 7√2 M1
when x = −√2, y = −7√2. A1

Answer (c): (√(2), 7√(2)) and (-√(2), -7√(2))

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Score: 0 / 10 (0 / 41 marks)

Circles and tangents Worksheets Next: Circle theorem proof →

Grade 8+ Quizzes

14 Topics

8A - Quad. sim. equations	8B - Transforming graphs y=f(x)
8C - Algebraic proof	8D - Completing the square
8E - Quadratic inequalities	8F - Quadratic sequences
8G - Velocity time graphs	8H - Circles and tangents
8J - Circle theorem proof	8K - Vector proof
8L - Probability equation questions	8M - Invariant points
8N - Spheres, cones and cylinders	8P - Volume of a frustum