Circle geometry is a constraint system. A diagram may look flexible, but equal radii, chord symmetry, tangent conditions and angle relationships restrict what is possible. Once the correct property is recognised, many apparently complicated diagrams collapse into short chains of reasoning.
This Secondary 3 Mathematics Learning Guide develops the core properties of circles through explanation, proof structure, worked examples and independent practice. It covers equal chords, perpendicular bisectors, tangents from an external point, angle in a semicircle, tangent-radius right angles, centre-circumference angle relationships, same-segment angles and supplementary opposite angles in cyclic quadrilaterals.
The official 2027 SEC G3 Mathematics syllabus K310 lists these symmetry and angle properties of circles within Geometry and Measurement. This guide stays with those stated relationships and uses original diagrams described in words rather than copied examination material.
Use the Secondary Mathematics Hub for the wider route. Inside this article, move through diagnostic · chords · tangents · angle properties · proof chains · practice · answers.
Begin With the Centre
The centre is often the hidden source of circle geometry. Every radius has equal length. Joining the centre to points on the circumference creates isosceles triangles. Joining the centre to a tangent point creates a right angle. Joining the centre to a chord can expose symmetry.
A useful first move is therefore to mark all known radii as equal and ask whether drawing an extra radius would reveal a triangle whose properties are already familiar.
A Six-Question Diagnostic
What angle is formed between a radius and a tangent at the point of contact? If AB is a diameter and C lies on the circle, what is angle ACB? If angle AOB at the centre subtends chord AB and is 120°, what is an angle ACB at the circumference subtending the same chord from the same segment? What can be said about tangents PA and PB from the same external point P? What is the sum of opposite angles in a cyclic quadrilateral? Finally, what happens when a line from the centre is perpendicular to a chord?
The answers are 90°; 90°; 60°; PA=PB; 180°; and the chord is bisected. These facts form a large part of the working vocabulary of circle geometry.
Equal Chords Are Equidistant From the Centre
If two chords in the same circle have equal lengths, their perpendicular distances from the centre are equal. Conversely, chords equidistant from the centre are equal in length.
The reason can be seen by dropping perpendiculars from the centre to each chord. Each perpendicular bisects its chord, and the resulting right triangles share equal hypotenuses because they are radii. Congruence then gives equal perpendicular distances when the half-chords match.
The Perpendicular Bisector of a Chord Passes Through the Centre
If M is the midpoint of chord AB and O is the centre, then OM is perpendicular to AB. Equivalently, the perpendicular bisector of any chord passes through the centre of the circle.
This property is useful when the centre is not shown. Constructing perpendicular bisectors of two different chords locates the centre at their intersection.
Worked Example 1: Find Half a Chord
A circle has radius 10 cm. Chord AB is 12 cm long. O is the centre and OM is perpendicular to AB. Find OM.
Because OM is perpendicular to chord AB from the centre, it bisects the chord. Hence AM=6 cm. In right triangle OMA, OA=10 and AM=6.
OM²=10²−6²=64, so OM=8 cm.
This example combines a circle property with Pythagoras. The circle property creates the right triangle; Pythagoras then completes the numerical work.
Worked Example 2: Compare Chords
Two chords in the same circle are 5 cm and 8 cm from the centre. Which chord is longer?
The chord closer to the centre is longer. Therefore the chord 5 cm from the centre is longer than the chord 8 cm from the centre.
This can be justified from right triangles: for a fixed radius r and perpendicular distance d from the centre, half-chord length is √(r²−d²). As d decreases, the half-chord increases.
A Radius Is Perpendicular to the Tangent at the Point of Contact
If line PT touches a circle at T and O is the centre, then OT is perpendicular to PT. The angle OTP is therefore 90°.
This right angle often unlocks a problem immediately because it creates a right triangle involving the centre, external point and tangent point.
Tangents From the Same External Point Have Equal Length
If PA and PB are tangents from external point P to a circle, then PA=PB. The centre line OP also bisects angle APB.
To see why, join OA, OB and OP. OA=OB because they are radii. OA and OB are perpendicular to the tangents. The two right triangles OAP and OBP share hypotenuse OP and have equal radius legs, so they are congruent. Hence PA=PB and the angles at P are equal.
Worked Example 3: Tangent Length
From external point P, tangents PA and PB touch a circle. If PA=13 cm, find PB.
By the equal-tangents property, PB=13 cm.
The point of the problem is not calculation. It is recognising that both segments are tangents from the same external point.
Worked Example 4: Radius, Tangent and External Distance
A circle has centre O and radius 9 cm. P is outside the circle and PT is tangent at T. If OP=15 cm, find PT.
OT is perpendicular to PT, so triangle OPT is right-angled at T. Therefore PT²=15²−9²=225−81=144.
Hence PT=12 cm.
Angle in a Semicircle Is 90°
If AB is a diameter and C is any other point on the circle, then angle ACB is 90°. The diameter subtends a right angle at the circumference.
This can transform an unfamiliar diagram into a right-triangle problem. If a line joining two points on the circle is known to pass through the centre, mark it as a diameter and look for the right angle it subtends.
Worked Example 5: Diameter Creates a Right Triangle
AB is a diameter of length 20 cm and C lies on the circle. AC=12 cm. Find BC.
Angle ACB=90°. Therefore AB is the hypotenuse of right triangle ACB.
BC²=20²−12²=400−144=256, so BC=16 cm.
The Angle at the Centre Is Twice the Angle at the Circumference
If angle AOB at the centre and angle ACB at the circumference subtend the same chord AB on the same arc, then angle AOB = 2 angle ACB.
Thus a 100° central angle corresponds to a 50° circumference angle standing on the same arc. The relationship works because both angles are tied to the same endpoints A and B.
Worked Example 6: Centre to Circumference
Angle AOB=136°. Point C lies on the opposite arc from AB and angle ACB subtends chord AB. Find angle ACB.
Angle at the centre is twice the angle at the circumference on the same arc, so angle ACB=136°/2=68°.
Angles in the Same Segment Are Equal
If points C and D lie on the same side of chord AB, then angles ACB and ADB subtending the same chord AB are equal.
This is a powerful identification tool. The two angles may be far apart in the diagram, but if they stand on the same chord and lie in the same segment, they are linked.
Worked Example 7: Same Segment
Chord AB is viewed from points C and D in the same segment. If angle ACB=47°, find angle ADB.
Angles in the same segment are equal, so angle ADB=47°.
Opposite Angles of a Cyclic Quadrilateral Are Supplementary
If A, B, C and D lie on the same circle, then ABCD is cyclic and opposite angles satisfy angle A + angle C = 180° and angle B + angle D = 180°.
This property works in reverse as a useful test: if a quadrilateral has a pair of opposite angles summing to 180°, it can be recognised as cyclic under the usual geometric conditions.
Worked Example 8: Cyclic Quadrilateral
ABCD is cyclic and angle ABC=112°. Find angle ADC.
Opposite angles are supplementary, so angle ADC=180°−112°=68°.
Circle Proof Is Usually a Chain of Small Reasons
A strong geometry proof does not merely write an angle value. It states why the value follows. Typical reasons include radius-radius equality, angle in a semicircle, tangent-radius perpendicularity, equal tangents, angles in the same segment, angle at centre twice angle at circumference, supplementary opposite angles and ordinary triangle angle facts.
The reasoning chain should be local: identify one new fact, justify it, then use it to obtain the next. Large jumps make proofs hard to verify and easy to break.
Worked Example 9: Tangents and an Isosceles Triangle
PA and PB are tangents from external point P. If angle APB=50°, find angles APO and OPB, where O is the centre.
The line OP bisects the angle between the two tangents. Therefore angle APO=angle OPB=50°/2=25°.
Alternatively, congruent right triangles OAP and OBP justify the angle-bisector result.
Worked Example 10: From Centre Angle to Cyclic Angle
A and B lie on a circle with centre O. Angle AOB=104°. Points C and D lie on the same arc opposite chord AB. Find angles ACB and ADB.
Each circumference angle subtending AB is half the central angle: 104°/2=52°. Therefore angle ACB=angle ADB=52°.
The equality can also be stated directly using the same-segment property once one of the two angles is known.
Worked Example 11: A Multi-Step Angle Chase
AB is a diameter. C and D are points on the circle. Angle CAB=28°. Find angle CBA, then explain one angle elsewhere on the circle that must also be 28°.
Since AB is a diameter, angle ACB=90°. In triangle ABC, angle CBA=180°−90°−28°=62°.
Any angle CDB standing on the same chord CB and lying in the same segment as angle CAB equals 28°. The exact named point D must be positioned so both angles subtend chord CB from the same segment.
Do Not Trust the Drawing’s Appearance
A chord that looks like a diameter may not pass through the centre. A quadrilateral that looks cyclic may not have all vertices on the circle. A line that looks tangent may not be stated as tangent.
Use only marked or logically deduced information. Geometry diagrams communicate structure, but they are not necessarily drawn to scale.
Four Common Errors
Using a circle property without matching the same chord: repair by naming the endpoints of the chord each angle subtends.
Assuming any line through a tangent point is perpendicular: only the radius to the point of contact is guaranteed perpendicular to the tangent.
Calling a quadrilateral cyclic because it looks inscribed: repair by confirming all four vertices lie on the circle or by proving a valid cyclic condition.
Giving an angle without a reason: repair by writing the property beside the calculation. Geometry marks often depend on the reasoning chain, not only the final number.
Independent Practice
1. A circle has radius 13 cm and a chord 10 cm long. The perpendicular from the centre meets the chord at M. Find the distance from the centre to M.
2. Two chords are equal. One is 6 cm from the centre. How far is the other from the centre?
3. From external point P, PA and PB are tangents. PA=17 cm. Find PB.
4. Radius OT=8 cm and OP=17 cm, with PT tangent at T. Find PT.
5. AB is a diameter of 26 cm and AC=10 cm. C lies on the circle. Find BC.
6. Angle AOB=150°. Find angle ACB subtending the same chord from the circumference.
7. Angles ACB and ADB lie in the same segment and ACB=39°. Find ADB.
8. ABCD is cyclic. Angle ABC=124°. Find angle ADC.
9. PA and PB are tangents and angle APB=64°. Find the two angles made by OP if OP bisects angle APB.
10. Explain why a radius drawn to a tangent point creates a right triangle.
11. A central angle is 88°. Find the corresponding circumference angle on the same arc.
12. A cyclic quadrilateral has angle A=73°. Find opposite angle C.
13. AB is a diameter and angle BAC=34°. Find angle ABC.
14. Explain why two equal chords are equally distant from the centre.
15. State one way to locate the centre of an unknown circle using chords.
Explained Answers
1. Half-chord=5. Distance=√(13²−5²)=√144=12 cm.
2. 6 cm, because equal chords in the same circle are equidistant from the centre.
3. 17 cm.
4. PT=√(17²−8²)=√225=15 cm.
5. Angle ACB=90°. BC=√(26²−10²)=√576=24 cm.
6. 75°.
7. 39°.
8. 180°−124°=56°.
9. 32° and 32°.
10. A radius is perpendicular to the tangent at the point of contact, so the angle there is 90°.
11. 44°.
12. 107°.
13. Angle ACB=90°, so angle ABC=56°.
14. Perpendiculars from the centre bisect equal chords, producing congruent right triangles with equal radii and equal half-chords; the perpendicular distances are therefore equal.
15. Draw two chords and construct the perpendicular bisector of each. Their intersection is the centre.
A Reliable Circle-Geometry Workflow
Mark the centre, radii, diameters, chords and tangent points. Identify the chord or arc associated with each angle. Add useful radii when they create isosceles or right triangles. Apply one property at a time and write the reason. Then check that the final angle fits triangle sums, cyclic sums and the visual structure.
Teacher and Parent Prompts
Ask “Which chord does this angle stand on?” and “What radius could we draw?” before offering a theorem. Ask the learner to justify every new equality or right angle. This makes geometry a chain of reasons rather than a list of memorised circle rules.
Continue the Secondary 3 Learning Route
Continue with Arc Length, Sector Area, Radians and Composite Mensuration for measurement inside circles, Congruence, Bisectors and Geometrical Construction for proof and construction, and Ratio, Proportion, Percentage, Rate and Speed in Real Contexts for multiplicative reasoning.
Circle geometry becomes secure when the learner can name the exact constraint that makes each step inevitable. Return to the Secondary Mathematics Hub.
