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Circles and tangents

Free Printable Circles and tangents PDF Worksheets, Revision Guide & GCSE Past Paper Questions (Grade 8+)

Download free circles and tangents worksheets (PDF), GCSE maths past paper questions and worked solutions for Circles and tangents (8H). This is a Grade 8+ algebra topic from the GCSE questions by topic collection for Foundation and Higher tier GCSE Maths (Edexcel, AQA, OCR and IGCSE).

The free PDF worksheets table below brings together printable question booklets (Qs) and worked answers or mark schemes (Ans) from Maths Genie, Corbett Maths, Maths Made Easy, Naiker Maths, 1st Class Maths, Metatutor, GCSE Maths Rocket, E Classroom and PMT, along with past paper question booklets.

Below the worksheet links you can read our circles and tangents revision notes and worked examples with diagrams, try real GCSE past paper circles and tangents questions with step-by-step solutions, or download a PDF of all circles and tangents worksheet links.

Free Circles and tangents PDF Worksheets & Answers

Grade 8+ • Topic 8H
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Grade 8+ topics
8A - Quad. sim. equations8B - Transforming graphs y=f(x)
8C - Algebraic proof8D - Completing the square
8E - Quadratic inequalities8F - Quadratic sequences
8G - Velocity time graphs8H - Circles and tangents
8J - Circle theorem proof8K - Vector proof
8L - Probability equation questions8M - Invariant points
8N - Spheres, cones and cylinders8P - Volume of a frustum
GCSE Maths Grade 8+ Revision Notes & Method

Guide to Circles and tangents: Explanation, Diagrams & Worked Examples

Download Printable One Page Guide (PDF)

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 mr = y1x1; (2) find the perpendicular tangent gradient mt = −1mr; (3) substitute mt and (x1, y1) into y − y1 = mt(x − x1).

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

O P(3, −4) Tangent

Solution:

Gradient of radius OP: mr = −4 − 03 − 0 = −43
Gradient of tangent (perpendicular): mt = +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: mr = 42 = 2 ⇒ mt = −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)
Scanned GCSE Exam Questions & Worked Solutions

GCSE Past Paper Circles and tangents Questions & Answers

Past Paper Example 1 AQA Jun 2017 1H Q27 4 marks
Full Paper (PDF) • Mark Scheme (PDF)
P(-1, 4) is a point on a circle, centre O
xyOPNot drawnaccurately
Work out the equation of the tangent to the circle at P.
Give your answer in the form y = mx + c
Worked Solution & Mark Scheme (4 marks)
Gradient of radius OP from (0, 0) to (-1, 4): M1
gradOP = 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
xyOPgrad = -4grad tangent = 1/4Not drawnaccurately
Answer: Answer y = 14x + 174
Past Paper Example 2 OCR Nov 2023 6H Q18 4 marks
Full Paper (PDF) • Mark Scheme (PDF)
A, B and C are points on the circumference of a circle.
ABCD8 cm58°45°Not to scale
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 & Mark Scheme (4 marks)
(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
Past Paper Example 3 Specimen Set 1 2H Q23 3 marks
Full Paper (PDF) • Mark Scheme (PDF)
Here is a circle, centre O, and the tangent to the circle at the point P(4, 3) on the circle.
P(4, 3)5– 5– 55Oxy
Find an equation of the tangent at the point P.
Worked Solution & Mark Scheme (3 marks)
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

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