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IGCSE · Practice questions · Circle theorems I: semicircles and tangents

Find the right angle.

Six original Cambridge-style questions on the angle in a semicircle and the tangent-radius theorem, with reasons for every step.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Spot the right angle, then use the triangle.

01
Recall
[2 marks]

Complete each statement. (a) The angle in a semicircle is ... (b) The angle between a tangent and a radius at the point of contact is ...

  • (a) \(90^\circ\) ✓
  • (b) \(90^\circ\) ✓
02
Calculation
[3 marks]

AB is a diameter of a circle and C lies on the circle. Angle BAC \(= 3x\) and angle ABC \(= 2x\). Find \(x\) and angle BAC.

  • Angle ACB \(= 90^\circ\) (angle in a semicircle). ✓
  • \(3x + 2x + 90 = 180\), so \(5x = 90\) and \(x = 18\). ✓
  • Angle BAC \(= 3 \times 18 = 54^\circ\). ✓
03
Calculation
[3 marks]

A circle has centre O and radius 8 cm. PT is a tangent to the circle at T and OP = 17 cm. Find PT.

  • Angle OTP \(= 90^\circ\) (angle between tangent and radius). ✓
  • \(PT^2 = 17^2 - 8^2 = 289 - 64 = 225\). ✓
  • PT = 15 cm. ✓
04
Analysis
[4 marks]

A circle has centre O. AB is a diameter and C is a point on the circle with angle OCB \(= 63^\circ\).

Find (a) angle OBC (b) angle BOC (c) angle OCA. Give a reason for each answer.

  • (a) OB = OC (radii), so triangle OBC is isosceles: angle OBC = angle OCB \(= 63^\circ\). ✓
  • (b) Angle BOC \(= 180^\circ - 63^\circ - 63^\circ = 54^\circ\) (angles in a triangle). ✓
  • (c) Angle ACB \(= 90^\circ\) (angle in a semicircle), so angle OCA \(= 90^\circ - 63^\circ = 27^\circ\). ✓
05
Proof
[4 marks]

AB is a diameter of a circle with centre O, and C is any point on the circle. Let angle OAC \(= a\) and angle OBC \(= b\).

Prove that angle ACB \(= 90^\circ\).

  • OA = OC (radii), so triangle OAC is isosceles and angle OCA \(= a\). ✓
  • OB = OC (radii), so triangle OBC is isosceles and angle OCB \(= b\). ✓
  • Angle ACB \(= a + b\). The angles of triangle ABC are \(a\), \(b\) and \(a + b\), so \(2a + 2b = 180^\circ\). ✓
  • Therefore \(a + b = 90^\circ\), so angle ACB \(= 90^\circ\). ✓
06
Analysis
[5 marks]

A satellite S is 8000 km from the centre O of a planet of radius 6400 km. The line of sight from S just touches the planet's surface at T.

(a) Explain why angle OTS \(= 90^\circ\). (b) Find the distance ST. (c) Find angle TOS, giving your answer to 1 decimal place.

  • (a) ST is a tangent to the circle and OT is a radius at the point of contact; the angle between a tangent and a radius is \(90^\circ\). ✓
  • (b) \(ST^2 = 8000^2 - 6400^2 = 64\,000\,000 - 40\,960\,000 = 23\,040\,000\). ✓
  • \(ST = \sqrt{23\,040\,000} = 4800\) km. ✓
  • (c) \(\cos(\text{TOS}) = \dfrac{OT}{OS} = \dfrac{6400}{8000} = 0.8\), so angle TOS \(= 36.9^\circ\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.