Secondary Mathematics · Worked Repair Guide 48
Many circle proofs are really symmetry arguments in disguise. Equal radii create isosceles triangles. A perpendicular from the centre to a chord creates two matching right triangles. Two tangents from the same external point produce another pair of matching right triangles. Once those congruent structures are visible, equal lengths and equal angles follow naturally.
This guide develops one central habit: draw the centre line that reveals the symmetry. A chord problem often becomes clear after joining the centre to the chord endpoints. A two-tangent problem often becomes clear after joining the external point to the centre.
All configurations below are original teaching constructions. This page is a narrow proof-and-symmetry companion to Circle Geometry — Arcs, Sectors, Chords and Tangents, not a replacement for that broader owner.
1. Equal radii create isosceles triangles
If O is the centre and A,B lie on the circle, then OA=OB because both are radii.
Therefore triangle OAB is isosceles and ∠OAB=∠OBA.
Entry check: if central angle AOB=80°, the two base angles are(180°−80°)/2=50°.
2. The perpendicular from the centre to a chord bisects the chord
Let OM be perpendicular to chord AB at M.
OA=OB, OM is common and both triangles OMA and OMB are right-angled at M.
Thus the two right triangles are congruent, so AM=MB.
The perpendicular centre line is therefore also a bisector of the chord.
3. The line from the centre to the midpoint of a chord is perpendicular to the chord
Now suppose M is the midpoint of chord AB, so AM=MB.
OA=OB and OM is common. Triangles OMA and OMB are congruent by SSS.
Therefore ∠OMA=∠OMB. These adjacent angles form a straight line, so each is90°.
Hence OM⊥AB.
4. Equal chords are equidistant from the centre
Suppose chords AB and CD have equal length. Draw perpendiculars OM and ON from the centre to the two chords.
The perpendiculars bisect the chords, so AM=CN when AB=CD.
OA=OC are radii. In right triangles OMA and ONC, equal hypotenuses and equal half-chords force OM=ON.
Thus equal chords lie at equal perpendicular distances from the centre.
5. Chords equidistant from the centre are equal
The converse also holds: if perpendicular distances OM and ON from the centre to chords AB and CD are equal, then the chords are equal.
Use right triangles with equal radii and equal perpendicular distances to prove equal half-chords, then double.
This gives a two-way relationship between chord length and distance from the centre.
6. Longer chords lie closer to the centre
In a circle of fixed radius r, a chord at perpendicular distance d from the centre has half-length √(r²−d²).
As d decreases, the half-chord increases.
Therefore among two chords in the same circle, the longer chord is closer to the centre.
The diameter is the limiting case: distance0 from the centre and maximum chord length2r.
7. Work a chord-distance calculation
A circle has radius13 cm and chord length10 cm.
The perpendicular centre line bisects the chord into5 cm halves.
Distance from centre to chord=√(13²−5²)=√144=12 cm.
The triangle is the 5-12-13 right triangle.
8. A tangent is perpendicular to the radius at the point of contact
If PT is tangent to a circle with centre O at T, then OT⊥PT.
This creates a right triangle whenever external point P is joined to O.
If OP=17 and radius OT=8, tangent length PT=√(17²−8²)=15.
The right angle is guaranteed by the tangent-radius theorem, not by the drawing.
9. Tangents from the same external point are equal
Let PA and PB be tangents from external point P, touching at A and B.
OA=OB are radii, OP is common, and ∠OAP=∠OBP=90°.
The right triangles OAP and OBP are congruent, so PA=PB.
This equality belongs specifically to tangents from the same external point.
10. The centre line bisects the angle between two tangents
From the same congruent triangles OAP and OBP, ∠APO=∠OPB.
Thus OP bisects angle APB.
The centre-to-external-point line is the symmetry axis of the two-tangent configuration.
It also bisects chord AB joining the tangent points.
11. The line joining tangent points is perpendicular to the centre line
In the symmetric two-tangent configuration, A and B are mirror-positioned around line OP.
Therefore OP is the perpendicular bisector of chord AB.
This can also be proved because OA=OB and PA=PB, so both O and P are equidistant from A and B; the line through two such points is the perpendicular bisector of AB.
12. The angle between two tangents is linked to the central angle
In quadrilateral OAPB, angles at A and B are90°.
Therefore ∠AOB+∠APB+90°+90°=360°.
Hence ∠APB=180°−∠AOB.
If central angle AOB=124°, the angle between tangents is56°.
13. Equal chords subtend equal central angles
If chord AB=CD, then triangles OAB and OCD have OA=OB=OC=OD as radii and AB=CD.
Thus the triangles are congruent by SSS, giving ∠AOB=∠COD.
Equal chords therefore correspond to equal central angles and equal minor arcs in the same circle.
14. Equal central angles produce equal chords
The converse follows from SAS: equal radii surround equal central angles.
So if ∠AOB=∠COD, then AB=CD.
This connects angular symmetry to length symmetry.
15. Chord length can be calculated from centre distance
In a circle radius10, a chord lies6 units from the centre.
Half-chord=√(10²−6²)=8, so full chord=16.
The centre distance must be perpendicular to the chord for this right-triangle construction.
16. Centre-line symmetry supports proof chains
A typical proof may move through several justified stages:
- radii are equal;
- tangent radii are perpendicular;
- right triangles are congruent;
- tangent lengths are equal;
- corresponding angles are equal;
- the centre line bisects an angle or chord.
The conclusion becomes reliable because every step has a stated reason.
17. Capstone: two tangents and the chord of contact
A circle has centre O and radius5. From external point P, OP=13 and tangents PA and PB touch the circle.
PA=PB=√(13²−5²)=12.
In right triangle OAP, sin∠APO=OA/OP=5/13, so ∠APO≈22.6°.
Since OP bisects ∠APB, the full tangent angle is about45.2°.
Therefore central angle AOB≈134.8°, agreeing with the supplementary relationship ∠APB+∠AOB=180°.
18. Independent practice
- If OA and OB are radii, what can be said about their lengths?
- If ∠AOB=80°, find base angles in isosceles triangle OAB.
- What does a perpendicular from the centre do to a chord?
- If a line from the centre meets a chord at its midpoint, what angle does it make with the chord?
- What can be said about equal chords and their distances from the centre?
- Which lies closer to the centre: the longer or shorter of two chords?
- A circle radius13 has chord length10. Find centre-to-chord distance.
- A circle radius10 has chord distance6 from centre. Find chord length.
- If PT is tangent at T, what is ∠OTP?
- If OP=17 and OT=8, find PT.
- If PA and PB are tangents from P and PA=9, find PB.
- What line bisects ∠APB in the two-tangent configuration?
- If ∠AOB=124°, find the angle between tangents at A and B.
- If equal chords AB and CD lie in one circle, compare ∠AOB and ∠COD.
- If equal central angles are given, compare their chords.
- Why does OP bisect chord AB joining the tangent points?
- If tangent angle APB=70°, find central angle AOB.
- In the 5-12-13 tangent configuration, find PA.
- Find half of the tangent angle in the capstone approximately.
- Explain the recurring role of congruent right triangles in chord and tangent proofs.
19. Worked answers
1. OA=OB.
2. 50° each.
3. It bisects the chord.
4. 90°.
5. They are equidistant from the centre.
6. The longer chord.
7. 12.
8. 16.
9. 90°.
10. 15.
11. 9.
12. OP.
13. 56°.
14. They are equal.
15. The chords are equal.
16. The configuration is symmetric: O and P are both equidistant from A and B, so OP is the perpendicular bisector of AB.
17. 110°.
18. 12.
19. Approximately22.6°.
20. Equal radii, common sides and tangent right angles repeatedly create congruent triangle pairs, allowing length and angle symmetry to be transferred.
20. Diagnose circle-symmetry errors by drawing the centre line
Common failures include using a non-perpendicular centre distance in a chord calculation, assuming unequal chords are equidistant from the centre, applying equal-tangent results to segments from different external points, or missing the symmetry line OP in a two-tangent configuration.
A useful repair note says “join the centre”, “drop the perpendicular”, “equal chords ↔ equal centre distances”, or “two tangents → congruent right triangles”.
21. Continue through the BTT learning routes
Return to the BTT Mathematics Hub or BTT Mathematical Lab. Use Circle Geometry, Congruence Tests and Pythagoras Converse for neighbouring proof tools.
Within Batch12, continue to Special Quadrilaterals or Data Collection, Frequency Tables and Dot Diagrams.
22. Sources and scope
The chord and tangent configurations are original teaching material. The page uses standard Euclidean circle theorems and congruence reasoning; theorem names and required proof style should be matched to the learner’s current syllabus.