Lemmara logoLemmaraWorked solution

0606/11 May/Jun 2025, Question 3: Circles Meeting Lines, and Tangents

Cambridge IGCSE Additional Mathematics 0606 · Paper 11 (Non-calculator) · Non-calculator · 6 marks

اقرأ هذا الحل مشروحاً بالعربي ←

The question, in short

A circle has equation (x − 4)² + (y + 3)² = 5 and point A is (3, −1). Show A is on the circle, find B given AB is a diameter, and find the tangent at A.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 1 mark

Show that the point (3, −1) satisfies the circle equation, confirming it lies on the circumference.

  1. 1Substitute x = 3, y = −1: (3 − 4)² + (−1 + 3)²
  2. 2= 1 + 4 = 5, which equals the right-hand side
  3. 3So A lies on the circumference

Answer(3 − 4)² + (−1 + 3)² = 1 + 4 = 5, so A lies on the circle (shown)

Part (b) · 2 marks

Using AB as a diameter, find the coordinates of the opposite endpoint B.

  1. 1Centre C = (4, −3)
  2. 2C is the midpoint of AB: (3 + x)/2 = 4 and (−1 + y)/2 = −3
  3. 3x = 5, y = −5

Answer(5, −5)

Part (c) · 3 marks

Find the equation of the tangent line to the circle at point A.

  1. 1Gradient of radius CA = (−3 + 1)/(4 − 3) = −2
  2. 2Gradient of tangent = −1/(−2) = 1/2
  3. 3y + 1 = (1/2)(x − 3)

Answery + 1 = (1/2)(x − 3)

Stuck on a step? Read the free lesson: Circles Meeting Lines, and Tangents

Practise more circles meeting lines, and tangents questions

Real 0606 past-paper questions, marked on your working like the exam, with a worked solution for every part.

Practise free on Lemmara

More past-paper questions on circles meeting lines, and tangents

Also searched as: 0606/11/M/J/25 · 0606_s25_qp_11 · 0606 s25 11 · question 3.
Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education. The worked solutions are Lemmara’s own.