Worked example · Easy
Halve the chord
A perpendicular from the centre meets a chord of length \(22\text{ cm}\) at \(M\). Find each half.
The perpendicular bisects the chord: \(22/2=\boxed{11\text{ cm}}\).
Chapter 38 · Completely rewritten
Circle theorems
Source-matched teaching notes, visual explanations, graduated worked examples and independent examination-style practice.
What you will learn
Use the key facts confidently, select an appropriate method, communicate each working step, and check the reasonableness of the final result.
- Symmetry properties of circles
- Angles in a circle
- The alternate segment theorem
1
Subtopic 1
Symmetry properties of circles
A chord joins two points on a circle. The perpendicular from the centre to a chord bisects the chord. Equal chords are equally distant from the centre. These properties follow from equal radii and congruent right triangles.
Tangents and radii
A tangent touches the circle once. At the touching point its radius is perpendicular to it. The two tangent segments from the same external point are equal in length. Radius, perpendicular distance and half-chord often form a right triangle, so use Pythagoras.
Worked examples: easy → medium → exam-style
Study the reason for each step; in an examination the method and relevant geometric explanation can earn marks even if a later calculation goes wrong.
Worked example · Easy
Equal tangents
Two tangents \(PA\) and \(PB\) start at P. Given \(PA=13\text{ cm}\), find \(PB\).
Tangents from the same point are equal: \(\boxed{PB=13\text{ cm}}\).
Worked example · Medium
Chord distance
A circle has radius \(10\text{ cm}\) and chord length \(12\text{ cm}\). Find its perpendicular distance from the centre.
Half-chord \(=6\text{ cm}\); by Pythagoras, distance \(=\sqrt{10^2-6^2}=\sqrt{64}=\boxed{8\text{ cm}}\).
Worked example · Medium
Tangent length
Point P is \(17\text{ cm}\) from the centre of a circle of radius \(8\text{ cm}\). Find tangent length.
Radius to contact makes \(90^\circ\). Length \(=\sqrt{17^2-8^2}=\sqrt{225}=\boxed{15\text{ cm}}\).
Worked example · Exam-style
An algebraic tangent pair
Tangents PA and PB have lengths \((4x+5)\) and \((6x-9)\text{ cm}\). Find \(x\) and both lengths.
Equal tangents: \(4x+5=6x-9\). Then \(2x=14\), \(x=7\). Each length \(=4(7)+5=\boxed{33\text{ cm}}\).
Worked example · Exam-style
Find a chord
A chord is \(5\text{ cm}\) from the centre of a circle of radius \(13\text{ cm}\). Find the full chord length.
Half-chord \(=\sqrt{13^2-5^2}=\sqrt{144}=12\). Full chord \(=\boxed{24\text{ cm}}\).
2
Subtopic 2
Angles in a circle
Before using a theorem, check that the angle is subtended by the correct arc. The angle at the centre is twice the angle at the circumference on the same arc. The angle in a semicircle is \(90^\circ\) because the central angle on a diameter is \(180^\circ\).
Circle angle rules
- Angles subtended by the same chord in the same segment are equal.
- Opposite angles of a cyclic quadrilateral add to \(180^\circ\).
- Angle between tangent and radius at touching point is \(90^\circ\).
- Two radii form an isosceles triangle; use equal base angles and the triangle sum.
Pay attention to reflex central angles and major/minor segments. If the diagram shows a chord that looks like a diameter, it is not a diameter unless stated or clearly marked as passing through the centre.
Worked examples: easy → medium → exam-style
Study the reason for each step; in an examination the method and relevant geometric explanation can earn marks even if a later calculation goes wrong.
Worked example · Easy
Central angle
An angle at the circumference is \(35^\circ\). Find the angle at the centre on the same arc.
Centre is double: \(2\times35=\boxed{70^\circ}\).
Worked example · Easy
Same segment
Angles subtended by chord AB in the same segment are \(48^\circ\) and \(x\).
Angles in the same segment are equal: \(\boxed{x=48^\circ}\).
Worked example · Medium
Angle in semicircle
ABC is a triangle drawn in a semicircle with AB the diameter. Given \(\angle A=38^\circ\), find \(\angle B\).
\(\angle C=90^\circ\). Hence \(\angle B=180-90-38=\boxed{52^\circ}\).
Worked example · Medium
Cyclic quadrilateral
Opposite angles of a cyclic quadrilateral are \((3x+12)^\circ\) and \((2x+18)^\circ\). Find \(x\).
Opposite angles sum to \(180\): \(5x+30=180\); \(\boxed{x=30}\).
Worked example · Medium
Isosceles radii
In triangle OAB, \(OA=OB\), and central angle \(\angle AOB=104^\circ\). Find each base angle.
\((180-104)/2=\boxed{38^\circ}\).
Worked example · Exam-style
Reflex central angle
The minor central angle subtended by AB is \(140^\circ\). Find the angle at the circumference on the other arc, subtended by the reflex arc AB.
Reflex central angle \(=360-140=220^\circ\). Corresponding circumference angle \(=220/2=\boxed{110^\circ}\).
Worked example · Exam-style
Two-step cyclic problem
A cyclic quadrilateral has \(\angle A=118^\circ\). An exterior angle adjacent to \(\angle B\) is \(74^\circ\). Find \(\angle C\) and \(\angle B\).
Opposite cyclic angles \(A+C=180\), so \(C=62^\circ\). Straight line at B: \(B=180-74=106^\circ\). Answers: \(\boxed{62^\circ,106^\circ}\).
3
Subtopic 3
The alternate segment theorem
The alternate segment theorem states that the angle between a tangent and a chord through the point of contact equals the angle in the opposite segment of the circle subtended by that chord.
How to apply it
Identify the point of tangency, trace the chord, and locate the inscribed angle on the other side of that chord. Mark the pair equal. In multi-step problems, combine it with the \(90^\circ\) radius–tangent angle, triangle sum and isosceles radii. State the theorem as your geometric reason.
The angle on the correct side of a tangent matters. Do not confuse the angle between tangent and chord with the one between tangent and radius.
Worked examples: easy → medium → exam-style
Study the reason for each step; in an examination the method and relevant geometric explanation can earn marks even if a later calculation goes wrong.
Worked example · Easy
Direct tangent–chord angle
At A, the tangent makes \(64^\circ\) with chord AB. Find the angle at C in the alternate segment subtended by AB.
By the alternate segment theorem, \(\angle ACB=\boxed{64^\circ}\).
Worked example · Easy
Reverse use
An angle at the circumference subtended by chord PQ is \(38^\circ\). Find the matching angle between PQ and the tangent at P.
By the alternate segment theorem, \(\boxed{38^\circ}\).
Worked example · Medium
Two tangent–chord angles
At one tangent contact A, angles between the tangent and two chords AB, AC on appropriate sides are \(52^\circ\) and \(73^\circ\). Find angles \(ACB\) and \(ABC\).
By the alternate segment theorem, \(\angle ACB=52^\circ\) and \(\angle ABC=73^\circ\).
Worked example · Medium
Complete the triangle
A triangle ABC lies on a circle with tangent at A. The alternate-segment angle to AB is \(46^\circ\), and \(\angle ABC=61^\circ\). Find \(\angle BAC\).
\(\angle ACB=46^\circ\); triangle sum gives \(A=180-46-61=\boxed{73^\circ}\).
Worked example · Exam-style
Algebra with alternate segment
Tangent–chord angle at A is \((3x+7)^\circ\), and its matching alternate-segment angle is \((5x-25)^\circ\). Find \(x\) and that angle.
\(3x+7=5x-25\); \(2x=32\); \(x=16\). Angle \(=3(16)+7=\boxed{55^\circ}\).
Worked example · Exam-style
Tangent plus isosceles triangle
Chord AB subtends \(56^\circ\) at point C, and \(OA=OB\). Find the angle between the tangent at A and AB, then the central angle AOB.
Alternate-segment angle \(=56^\circ\). Central angle on same minor arc \(=2\times56=\boxed{112^\circ}\).
★
Independent practice
Exercises with step-by-step Show Answer solutions
Questions in each subtopic progress from confidence-building to developed methods and original exam-style problems. The problems are original practice, not copied past-paper questions. Answers stay closed until selected.
Symmetry properties of circles
Work through the basics first; save the examination-style questions for last.
1Question 1Basic
A chord is \(18\text{ cm}\). A radius through its midpoint meets it at right angles. Find each half.
Show Answer
Step-by-step solution: \(18/2=\boxed{9\text{ cm}}\).
2Question 2Basic
From point P, tangent PA has length \(9\text{ cm}\). Find tangent PB to the same circle.
Show Answer
Step-by-step solution: Equal tangents from P: \(\boxed{9\text{ cm}}\).
3Question 3Basic
What angle does a radius make with a tangent at the point of contact?
Show Answer
Step-by-step solution: The radius is perpendicular to the tangent: \(\boxed{90^\circ}\).
4Question 4Developing
A circle has radius \(5\text{ cm}\). A chord is \(6\text{ cm}\) long. Find its distance from the centre.
Show Answer
Step-by-step solution: Half-chord \(3\), distance \(\sqrt{5^2-3^2}=\boxed{4\text{ cm}}\).
5Question 5Developing
A circle has radius \(15\text{ cm}\), chord distance \(9\text{ cm}\). Find chord length.
Show Answer
Step-by-step solution: Half-chord \(\sqrt{15^2-9^2}=12\); full chord \(\boxed{24\text{ cm}}\).
6Question 6Developing
Point P is \(10\text{ cm}\) from centre O of a circle radius \(6\text{ cm}\). Find the tangent from P.
Show Answer
Step-by-step solution: Right triangle: \(\sqrt{10^2-6^2}=\boxed{8\text{ cm}}\).
7Question 7Exam-style
Tangents from P are \((5x-2)\text{ cm}\) and \((3x+14)\text{ cm}\). Find \(x\).
Show Answer
Step-by-step solution: \(5x-2=3x+14\); \(2x=16\); \(\boxed{x=8}\).
8Question 8Exam-style
From a point P, \(OP=26\text{ cm}\), radius \(OA=10\text{ cm}\), and PA is tangent. Find PA.
Show Answer
Step-by-step solution: \(PA=\sqrt{26^2-10^2}=\sqrt{576}=\boxed{24\text{ cm}}\).
9Question 9Exam-style
Two chords in the same circle are equal. One lies \(7\text{ cm}\) from the centre. How far is the other?
Show Answer
Step-by-step solution: Equal chords are equally distant from the centre: \(\boxed{7\text{ cm}}\).
Angles in a circle
Work through the basics first; save the examination-style questions for last.
10Question 10Basic
A circumference angle is \(27^\circ\). Find the central angle subtending the same arc.
Show Answer
Step-by-step solution: \(2\times27=\boxed{54^\circ}\).
11Question 11Basic
An angle at the centre is \(118^\circ\). Find the angle at the circumference on the same arc.
Show Answer
Step-by-step solution: \(118/2=\boxed{59^\circ}\).
12Question 12Basic
A triangle is inscribed in a semicircle with its longest side as diameter. Find the opposite angle.
Show Answer
Step-by-step solution: Angle in a semicircle \(=\boxed{90^\circ}\).
13Question 13Basic
Two angles in the same segment subtend one chord. One is \(41^\circ\). Find the other.
Show Answer
Step-by-step solution: Equal angles in the same segment: \(\boxed{41^\circ}\).
14Question 14Developing
In a cyclic quadrilateral one angle is \(124^\circ\). Find its opposite.
Show Answer
Step-by-step solution: \(180-124=\boxed{56^\circ}\).
15Question 15Developing
A central angle is \(86^\circ\). OA and OB are radii. Find each base angle of OAB.
Show Answer
Step-by-step solution: \(OA=OB\); base angles \((180-86)/2=\boxed{47^\circ}\).
16Question 16Developing
In a triangle inscribed in a semicircle, one acute angle is \(31^\circ\). Find the other.
Show Answer
Step-by-step solution: Other acute angle \(=180-90-31=\boxed{59^\circ}\).
17Question 17Developing
Two opposite angles of a cyclic quadrilateral are \((x+18)^\circ\) and \((2x+12)^\circ\). Find \(x\).
Show Answer
Step-by-step solution: \(3x+30=180\), so \(\boxed{x=50}\).
18Question 18Exam-style
An arc subtends \(144^\circ\) at the centre. Find the circumference angle on the same arc.
Show Answer
Step-by-step solution: \(144/2=\boxed{72^\circ}\).
19Question 19Exam-style
A minor central angle measures \(100^\circ\). Find the angle at the circumference in the corresponding opposite (major) segment.
Show Answer
Step-by-step solution: The required angle subtends the minor arc, so \(100/2=\boxed{50^\circ}\). State which arc is subtended.
20Question 20Exam-style
A cyclic quadrilateral has opposite angles \((4x+5)^\circ\) and \((3x+28)^\circ\). Find \(x\).
Show Answer
Step-by-step solution: \(7x+33=180\), so \(7x=147\) and \(\boxed{x=21}\).
21Question 21Exam-style
At a point on a circle, a tangent and radius meet. The angle between the tangent and a chord is \(36^\circ\) on one side. Find the remaining angle between the radius and chord on that side.
Show Answer
Step-by-step solution: Tangent–radius is \(90^\circ\), so angle \(=90-36=\boxed{54^\circ}\).
The alternate segment theorem
Work through the basics first; save the examination-style questions for last.
22Question 22Basic
A tangent–chord angle is \(28^\circ\). Find its alternate-segment angle.
Show Answer
Step-by-step solution: Equal by alternate segment theorem: \(\boxed{28^\circ}\).
23Question 23Basic
An inscribed angle is \(67^\circ\). Find the corresponding angle between the tangent and chord.
Show Answer
Step-by-step solution: Equal by alternate segment theorem: \(\boxed{67^\circ}\).
24Question 24Basic
Tangent–chord angle is \(49^\circ\); name the theorem that makes the inscribed angle equal.
Show Answer
Step-by-step solution: The alternate segment theorem: inscribed angle \(=\boxed{49^\circ}\).
25Question 25Developing
A tangent–chord angle is \((2x+15)^\circ\), and the alternate angle is \(73^\circ\). Find \(x\).
Show Answer
Step-by-step solution: \(2x+15=73\); \(2x=58\); \(\boxed{x=29}\).
26Question 26Developing
Triangle ABC inscribed in a circle has a tangent at A. The tangent–AB angle is \(42^\circ\), and \(\angle ABC=76^\circ\). Find \(\angle BAC\).
Show Answer
Step-by-step solution: \(\angle ACB=42^\circ\). Then \(A=180-76-42=\boxed{62^\circ}\).
27Question 27Developing
Tangent makes \(35^\circ\) with a chord at the touching point; find the angle between the chord and radius on the same side.
Show Answer
Step-by-step solution: Radius–tangent \(90^\circ\). So chord–radius \(=90-35=\boxed{55^\circ}\).
28Question 28Exam-style
Alternate angles subtended by the same chord are \((5x-4)^\circ\) and \((2x+44)^\circ\). Find \(x\).
Show Answer
Step-by-step solution: Equate: \(5x-4=2x+44\); \(3x=48\); \(\boxed{x=16}\).
29Question 29Exam-style
In a circle tangent–chord angle is \(58^\circ\) and the other non-adjacent triangle angle is \(47^\circ\). Find the triangle angle at the tangent point.
Show Answer
Step-by-step solution: Alternate angle is \(58^\circ\). Triangle sum \(=180-58-47=\boxed{75^\circ}\).
30Question 30Exam-style
At a circle, a tangent meets chords AB and AC; the opposite angles in triangle ABC are \(45^\circ\) and \(75^\circ\). Find the third triangle angle.
Show Answer
Step-by-step solution: Angles in a triangle sum to \(180^\circ\), so \(180-45-75=\boxed{60^\circ}\). The tangent–chord angles match the opposite angles by the alternate segment theorem.
31Question 31Exam-style
Tangent at A makes \(68^\circ\) with chord AB. The central angle on the minor arc AB is required. Find it.
Show Answer
Step-by-step solution: Alternate segment angle \(ACB=68^\circ\); centre angle subtending same arc \(=2(68)=\boxed{136^\circ}\).
High-grade revision checklist
- Can I explain and correctly apply: Symmetry properties of circles?
- Can I explain and correctly apply: Angles in a circle?
- Can I explain and correctly apply: The alternate segment theorem?
- Have I shown reasons, units, and sufficient steps rather than just giving final numbers?
- Have I checked which quantity or angle the question actually asks for?