A circle has radius 17 cm and a chord 30 cm long. How far is the chord from the centre? There is no need to search for a special chord formula. The important move is to draw a perpendicular from the centre to the chord. Symmetry divides the chord into two equal pieces, and a familiar right triangle appears. The distance is 8 cm. The circle has not introduced an unrelated kind of Mathematics; it has organised ideas you already know into a new arrangement.
This guide develops circle geometry through lengths, perpendicular distances and symmetry. Its companion on circle angles and tangent reasoning handles angle chains. Here, the task is to identify which segments are equal, which triangle is right-angled, and which conditions justify those conclusions before calculating.
Secondary 2-to-upper-secondary bridge. Circle properties are listed in the Secondary Three/Four sections of the MOE G2 and G3 Mathematics syllabuses, including the circle sections on printed pages 20 and 30. This article gives a prepared Secondary 2 learner an optional, guided bridge; it does not present these theorems as compulsory Secondary 2 content for every subject level. Follow your school’s current sequence. All worked questions are original teaching examples.
Secondary Mathematics Hub · Batch 5. The other routes are scale drawings and constructions and algebraic fractions and restrictions.
Read by task: read the circle · chords and distances · compare chords · tangents · a connected investigation · practice with explained answers · teaching and independent return.
1. Read the circle as a collection of constraints
A circle consists of points at one fixed distance from a centre. That distance is the radius. The filled region inside is a disc. A chord joins two points on the circumference; a diameter is a chord passing through the centre. Every diameter has length twice the radius, but most chords are not diameters. A secant line cuts the circle at two points, whereas a tangent line meets it at one point.
These definitions tell you which information is available. If O is the centre and A and B lie on the circumference, OA = OB because both are radii of the same circle. If P is merely somewhere inside the circle, OP need not equal either radius. If AB passes near O but not through O, it is not a diameter. A useful sketch keeps these distinctions visible rather than smoothing them away.
What does distance from a point to a chord mean?
The distance from the centre to a chord is the perpendicular distance to its supporting line. It is not the length of a slanting segment drawn from the centre to a convenient endpoint. If M is the foot of the perpendicular from O to AB, then OM is the required distance. OA is a radius. AM is part of a chord. Naming all three roles prevents a formula from being filled with incompatible lengths.
Draw a circle, choose a chord AB that is not a diameter, mark the centre O and draw OM perpendicular to AB. Join OA and OB. You now have two right triangles sharing OM. This is the working picture for much of the guide. No printed diagram is needed: your own labelled sketch makes the given and derived information explicit.
A readiness check before introducing a theorem
Check three prerequisites. Can you explain why two radii of one circle are equal? Can you identify the hypotenuse from a right-angle marker rather than from the apparent size of a drawing? Can you use 8² + 15² = 17² to find any missing member of the triple? If one of these is unstable, use the earlier geometry and constraints guide before adding a longer circle chain.
There is also a language check. The statements OM is perpendicular to AB and M is the midpoint of AB do different jobs. Perpendicularity supplies a right angle. Midpoint supplies AM = MB. A circle theorem can connect these statements under suitable conditions, but they should not be treated as synonyms in every diagram.
2. Why the perpendicular from the centre bisects a chord
Let AB be a non-diameter chord and let OM be perpendicular to AB, with M on AB. Compare right triangles OMA and OMB. Their hypotenuses OA and OB are equal radii. Their leg OM is shared. The triangles are congruent by the right-angle, hypotenuse and side criterion. Consequently AM = MB. The perpendicular from the centre reaches the midpoint of the chord.
This argument also explains the symmetry: reflection across the line OM exchanges A and B while leaving O fixed. The standard chord-bisection relationship and its converse appear in Euclid, Book III, Proposition 3. In a school solution, the important part is to state the centre and perpendicular conditions, not simply write a theorem name beside an unexplained halving.
The converse has conditions too
Suppose M is the midpoint of a non-diameter chord AB. Then OA = OB, AM = MB and OM is shared, so triangles OMA and OMB are congruent by three matching sides. The equal angles at M also lie on a straight line. Each is therefore 90°, and OM is perpendicular to AB.
The non-diameter qualification avoids a collapsed construction: for a diameter, its midpoint is O, so the segment OM has zero length and cannot supply a distinct centre-to-midpoint direction. The perpendicular bisector of a diameter still passes through O, but it should be described as a line rather than as a non-zero segment joining O to itself. Boundary cases sharpen a theorem instead of weakening it.
Worked problem 1: the opening chord question
A circle has radius 17 cm and chord AB = 30 cm. Find the perpendicular distance from O to AB. Draw OM perpendicular to AB. Chord bisection gives AM = 15 cm. Triangle OMA is right-angled, with OA = 17 cm as its hypotenuse. Thus OM² = 17² − 15² = 289 − 225 = 64, and OM = 8 cm.
Two checks are available. The distance 8 cm is less than the radius, as it must be for a non-degenerate chord. Reconstructing the triangle gives 8² + 15² = 17². Using 30 instead of 15 inside the right triangle would produce a negative square; that would reveal a structural mistake, not a reason to change the theorem.
Worked problem 2: find the whole chord
A chord is 7 cm from the centre of a circle of radius 25 cm. Find its length. With M the perpendicular foot, AM² = 25² − 7² = 625 − 49 = 576. Hence AM = 24 cm. The complete chord is twice this value, so AB = 48 cm.
The last doubling is essential. Pythagoras found a side of one of the two right triangles, not automatically the quantity named in the question. The full chord is shorter than the 50 cm diameter, providing a second check. A final answer of 24 cm would reveal unfinished interpretation even though every arithmetic operation was correct.
Worked problem 3: reconstruct the radius
A chord is 40 cm long and 21 cm from the centre. Find the radius. Half the chord is 20 cm. The radius is the hypotenuse of a right triangle with legs 20 and 21, so r² = 20² + 21² = 400 + 441 = 841. Therefore r = 29 cm.
Compare all three worked problems. Sometimes the unknown is the hypotenuse; sometimes it is a leg; sometimes the requested result is twice a leg. The diagram remains similar, but the algebraic operation changes with the unknown’s role. Naming the roles is more dependable than memorising that chord questions always involve subtraction.
3. Equal chords, equal distances and a fixed radius
Let a circle have radius r, chord half-length a and perpendicular distance d from its centre. The right triangle gives a² + d² = r². With r fixed, equal half-chords give equal perpendicular distances, and equal perpendicular distances give equal half-chords. The full chords are equal exactly when their half-lengths are equal. This is the relationship formalised in Euclid, Book III, Proposition 14.
The phrase with r fixed is doing real work. Two chords in circles of different radii may be equal without being equally far from their respective centres. For example, a chord of length 12 in a circle of radius 10 lies 8 units from the centre. A chord of length 12 in a circle of radius 13 lies √133 units from the centre. Equal chord lengths alone do not overcome the different radii.
Worked problem 4: compare without measuring the drawing
Two chords in a circle of radius 15 cm lie 9 cm and 12 cm from the centre. Find their lengths. For the first, half-length = √(225 − 81) = 12 cm, so the chord is 24 cm. For the second, half-length = √(225 − 144) = 9 cm, so the chord is 18 cm. The chord closer to the centre is longer.
The same conclusion can be predicted before calculation. Increasing d leaves a smaller value for a² in a² = r² − d². Therefore a decreases. At d = 0 the chord is a diameter. As d approaches r, the chord length approaches zero. At d = r the two endpoints coincide, so the limiting contact is not an ordinary chord with two distinct endpoints.
An optional compact formula, derived rather than borrowed
The full chord length L satisfies L = 2√(r² − d²) for 0 ≤ d < r. This is simply the half-chord triangle written in one line. It is useful after the geometry is understood, not as a replacement for understanding. If you forget it, reconstruct the diagram, halve the chord and use Pythagoras.
The formula also provides an existence test. A claimed chord of length 31 cm cannot fit in a circle of radius 15 cm, because the largest chord is the 30 cm diameter. A claimed distance d greater than r would make the square root negative and indicates that the line misses the circle. A calculation must answer to the object’s geometry.
4. Tangents create a different right triangle
A tangent at A is perpendicular to the radius OA drawn to the point of contact. The contact point matters: a radius to some other point on the circle need not be perpendicular to that tangent. This standard relationship is stated in Euclid, Book III, Proposition 18.
Let P be outside the circle and PA a tangent. Joining OP creates right triangle OAP, right-angled at A. Now OP is the hypotenuse. OA is a radius and PA is a tangent segment. The longest side is not automatically the radius simply because the picture contains a circle. The right-angle marker decides the hypotenuse.
Worked problem 5: find a tangent length
OP = 20 cm, the radius OA = 12 cm, and PA is tangent at A. Find PA. Since OA is perpendicular to PA, PA² = OP² − OA² = 400 − 144 = 256. Therefore PA = 16 cm. The subtraction 20 − 12 = 8 does not find PA because the centre distance and radius are not consecutive pieces of the same straight segment along the tangent.
Check 12² + 16² = 20². Also check that P is outside: OP exceeds the radius. If OP were smaller than the radius, no real tangent segment could be drawn from P to the circle. That is a geometrical impossibility, not an invitation to ignore a negative value under a square root.
Two tangents from one external point have equal lengths
Suppose PA and PB are tangents from the same external point P, touching the circle at A and B. Triangles OAP and OBP are right-angled at their contact points. OA = OB because they are radii, and OP is their common hypotenuse. Right-triangle congruence gives PA = PB.
The shared external point is essential. Two tangent segments drawn from different external points need not have equal lengths. Nor does the conclusion say that PA equals the radius. It equates the two tangent lengths belonging to this particular pair of congruent triangles. A correct rule becomes unreliable when its conditions are silently dropped.
Worked problem 6: tangent equality produces algebra
PA and PB are tangents from P. Their lengths are (3x + 2) cm and (5x − 8) cm. Find x and both lengths. Set 3x + 2 = 5x − 8, giving 10 = 2x and x = 5. Each tangent is 17 cm. Substitute into both expressions; do not stop after finding x if the question asks for the lengths.
Now imagine the same expressions label one tangent and one arbitrary chord. The equation 3x + 2 = 5x − 8 would no longer be justified. Algebra solves the equation that geometry permits. It cannot create a missing equality simply because the expressions look convenient.
The centre line also bisects the tangent angle
The same congruent triangles give ∠APO = ∠OPB. Thus OP bisects the angle between PA and PB. They also give ∠AOP = ∠POB. One carefully established congruence supports several conclusions: equal tangent lengths, equal angles at P and equal angles at O. It is more useful to understand this common proof than to store each result as an unrelated fact.
For example, if ∠APB = 70°, each angle at P in the two right triangles is 35°. Each corresponding angle at O is 55°. These values agree with the triangle sums. The angle companion develops longer chains, while the present guide keeps the focus on how this symmetry helps with lengths.
5. A connected investigation: the chord between the contact points
Guided extension. Draw a circle of radius 15 cm with external point P such that OP = 25 cm. Draw its two tangents PA and PB. Join the contact points A and B, and let M be the intersection of AB with OP. The questions are to find the tangent length, OM and AB. This combines previously justified facts rather than introducing a new formula without explanation.
Both O and P are equidistant from A and B: OA = OB and PA = PB. Therefore they lie on the perpendicular bisector of AB. Since O and P are distinct, their joining line is that perpendicular bisector. Hence OP is perpendicular to AB and M is its midpoint. The construction guide explains this equidistance property in detail.
Solve the first triangle, then compare two triangles
Triangle OAP gives PA = √(25² − 15²) = 20 cm. To obtain OM, compare triangles OMA and OAP. They have equal angles at O because OM and OP point along the same line towards P, and each has a right angle. Therefore they are similar. The matching hypotenuses are OA in the smaller triangle and OP in the larger.
The corresponding sides next to the shared angle give OM/OA = OA/OP. Thus OM/15 = 15/25, so OM = 9 cm. Finally AM = √(15² − 9²) = 12 cm, and AB = 24 cm. The ratio can be checked: the small-to-large length factor is 15/25 = 3/5, and AM/PA = 12/20 = 3/5.
This is a useful problem because each result unlocks the next one. Right angles establish triangle structure. Congruence establishes symmetry. Similarity finds an internal distance. Pythagoras finds a half-chord. Doubling returns the full chord. The route is longer than any individual calculation, but each step has a clearly stated reason.
A second extension: infer a circle from a shallow segment
A chord has length 16 cm. At its midpoint, the perpendicular distance to the nearer arc is 4 cm. Find the radius. This 4 cm height is sometimes called the segment height or sagitta. Draw the radius through the chord midpoint M to the midpoint of the nearer arc. If the radius is r, the centre-to-chord distance is r − 4, and the half-chord is 8.
Pythagoras gives (r − 4)² + 8² = r². Expanding yields r² − 8r + 16 + 64 = r², so 8r = 80 and r = 10 cm. The centre-to-chord distance is 6 cm. Check 6² + 8² = 10² and check that 10 − 6 returns the specified 4 cm segment height.
The phrase nearer arc identifies the short cap and its geometry. Calling any convenient vertical distance 4 cm would not support the same equation. In a practical measurement setting, rounding of chord length and segment height would also affect the inferred radius. The example treats the stated lengths as exact mathematical data, not as a claim of unlimited measuring precision.
6. Practice: identify the right triangle before calculating
Make a labelled sketch for every geometric question. Do not start by memorising the worked numbers. Write the required quantity, mark any midpoint derived from a theorem, and label the hypotenuse. Questions 1–16 consolidate the core relationships in this bridge; questions 17–20 are guided extensions. All circles and lengths below use ordinary Euclidean geometry.
Questions 1–5: chords
1. A chord is 6 cm from the centre of a circle of radius 10 cm. Find its length. 2. A circle has radius 13 cm and chord length 24 cm. Find the perpendicular centre-to-chord distance. 3. A chord is 20 cm long and 24 cm from the centre. Find the radius. 4. Two chords of the same circle have equal lengths. One is 7 cm from the centre. How far is the other? 5. Explain why an arbitrary perpendicular drawn to a chord need not bisect it.
Questions 6–10: conditions and tangents
6. In a circle of radius 15 cm, compare chord lengths at distances 9 cm and 12 cm from the centre. 7. Can a circle of radius 8 cm have a chord 17 cm long? Explain. 8. P is outside a circle, PA is tangent at A, OP = 13 cm and OA = 5 cm. Find PA. 9. Two tangents from P have lengths (4x + 1) cm and (6x − 9) cm. Find x and the common length. 10. In triangle OAP from Question 8, name the hypotenuse and explain why.
Questions 11–15: reason rather than assume
11. Tangents PA and PB form an angle of 64° at P. Find ∠APO. 12. In the same configuration, find ∠AOP. 13. A tangent segment is 15 cm long and the radius at contact is 8 cm. Find the distance from the external point to the centre. 14. Explain why equal tangent segments from different external points cannot be assumed. 15. A non-diameter chord AB has midpoint M. O is the centre. State one triangle-congruence route that establishes OM perpendicular to AB.
Questions 16–20: changed conditions and extensions
16. A chord passes through the centre of a radius-11 cm circle. Find its length and centre-to-chord distance. 17. In the two-tangent configuration, radius OA = 10 cm and OP = 25 cm. If M is where OP meets contact chord AB, use similarity to find OM. 18. Find AB in Question 17, leaving an exact square-root answer. 19. A chord of length 12 cm has nearer-arc segment height 2 cm. Find the radius. 20. Explain how two non-parallel chords can be used to recover the centre of a drawn circle.
Explained answers 1–5
1. Half-chord = √(100 − 36) = 8 cm, so the full chord is 16 cm. 2. Half the chord is 12 cm. Distance = √(169 − 144) = 5 cm. 3. The radius is the hypotenuse: √(10² + 24²) = 26 cm. These three questions deliberately change which length is unknown.
4. The other distance is 7 cm, because the chords belong to the same circle. 5. The chord-bisection theorem requires the perpendicular to pass through the centre. A perpendicular line crossing near one endpoint is still perpendicular but does not generally divide the chord equally. The missing centre condition is the issue.
Explained answers 6–10
6. The lengths are 24 cm and 18 cm, respectively, from twice √(225 − d²). The nearer chord is longer. 7. No: the diameter is 16 cm, the maximum chord length. 8. PA = √(13² − 5²) = 12 cm, because the radius is perpendicular to the tangent.
9. Set 4x + 1 = 6x − 9. Then x = 5 and both tangents are 21 cm. 10. OP is the hypotenuse because it is opposite the right angle at A. The name of the point at which the right angle occurs is more reliable than the apparent longest line in a rough sketch.
Explained answers 11–15
11. OP bisects the angle between the tangents, so ∠APO = 32°. 12. Triangle OAP has a 90° angle at A and 32° at P, leaving 58° at O. 13. The centre distance is √(15² + 8²) = 17 cm. Here the unknown is the hypotenuse, so the squares are added.
14. The equal-tangent argument uses a common external point and a shared centre-distance hypotenuse. Different external points do not supply that shared triangle data. 15. OA = OB, AM = MB and OM is shared, so triangles OMA and OMB are congruent by SSS. Their equal angles at M sum to 180°, making both 90°.
Explained answers 16–20
16. The chord is a diameter of 22 cm, and its distance from the centre is 0 cm. 17. From similar triangles OMA and OAP, OM/10 = 10/25, so OM = 4 cm. 18. Half-chord = √(10² − 4²) = √84 = 2√21 cm, so AB = 4√21 cm, approximately 18.33 cm.
19. Let radius be r. The half-chord is 6 and centre distance r − 2. Then (r − 2)² + 36 = r², giving −4r + 40 = 0. Thus r = 10 cm. 20. Construct each chord’s perpendicular bisector. The centre lies on both because it is equidistant from each chord’s endpoints. Their intersection identifies the centre; check equal distances to several circumference points.
7. Turn a completed solution into a reusable method
For teaching, separate three tasks that are often rushed together. First ask the learner to add the useful lines without calculating. Next ask which equality or right angle follows and why. Only then insert the numerical lengths. This makes it possible to distinguish a diagram-reading problem from a Pythagoras calculation problem.
A short repair sequence can keep the radius fixed while moving a chord closer to and farther from the centre. Ask for a prediction before calculation. Then keep the chord length fixed and change the radius, testing whether the learner remembers the same-circle qualification. These are proposed teaching contrasts, not promises that a fixed number of questions will produce mastery.
Correct the first unsupported step
If a student uses the full chord in the right triangle, the repair is half-chord recognition. If the student identifies the right triangle but subtracts in the wrong direction, repair hypotenuse ownership. If two arbitrary tangents are set equal, repair the common-external-point condition. If an algebraic x is correct but the requested length is missing, repair the final substitution and interpretation.
An error record should capture the first unsupported line, the condition that was missing and a fresh return problem. For instance: I halved AB because a perpendicular touched it, but that perpendicular did not pass through O. The return question should include both a centre perpendicular and an unrelated perpendicular, requiring the student to say which one actually creates equal chord halves.
Parent and tutor prompts
Ask where the centre is, which lengths are radii, which right angle is justified, and whether the calculation has found half a chord or the whole chord. For tangent problems, ask whether the two tangents begin at the same external point. These questions invite an explanation without supplying the next numerical operation.
A useful independent return changes the diagram’s orientation and the location of the unknown. Rotate the chord, put its length into an algebraic expression or supply the tangent length instead of the centre distance. The learner should rebuild the same relationships rather than recognise the position of a familiar number. A correct answer plus a condition-based explanation is stronger evidence than an answer obtained by tracing a remembered layout.
8. Where this bridge leads next
Circle lengths connect several earlier capabilities: equality of radii, congruent triangles, perpendicular bisectors, Pythagoras, similarity and algebraic rearrangement. The circle’s symmetry is the reason those capabilities fit together. The next step is not to memorise every imaginable circle formula, but to become more deliberate about which relationship a particular configuration supports.
Continue with Circle Angles, Semicircles and Tangent Reasoning for intercepted arcs and angle chains. Use Scale Drawings and Construction to build the perpendicular-bisector relationships yourself. When algebraic restrictions become the difficulty, use Algebraic Fractions, Restrictions and Fractional Equations.
For the later-year treatment, the existing Secondary 3 Circle Properties guide remains a separate progression route. This preparatory guide does not replace that owner or claim that completing a bridge changes the learner’s school syllabus.
Read the conditions. Draw the useful radius. Establish the perpendicular or equality. Solve the resulting triangle. Return from the intermediate length to the quantity the question asked for.
Reference boundary. Curriculum placement refers to the linked MOE syllabus. Mathematical theorem references are the linked Euclid propositions in David Joyce’s academic edition at Clark University. The numerical questions, solution routes, counterexamples and teaching sequence in this guide are original. They are not official examination questions or a claim of endorsement by those institutions.
