I'm Up and Down, and Round and Round

Why This Matters

Look around and you will see circles everywhere. A bicycle wheel. A full moon. The ripples a raindrop makes when it falls on a puddle. The round roti on your plate. Humans have been fascinated by this shape for thousands of years. Some of the oldest cave paintings in India, at Gudahandi in Odisha, are full of circles drawn beside triangles and squares.

What makes a circle so special? It is the most perfectly balanced shape there is. Spin a wheel and it looks exactly the same at every moment. Fold a round paper in half along any line through its middle and the two halves match perfectly. No other shape is this smooth and even.

In this chapter we go past just looking at circles. We find their hidden rules. Why does a longer chord sit closer to the centre? Why is a triangle drawn inside a semicircle always a right-angled triangle? Why do the opposite corners of a four-sided shape drawn inside a circle always add up to 180°? These are not random facts to memorise. Each one has a clean reason, and we will prove every one of them, step by step, with a picture. Once you see why, you will never forget it.

The Big Idea

The Big Idea: A circle is simply all the points that are the same distance from one fixed point. That fixed point is the centre, and that fixed distance is the radius. This one idea — “same distance from the centre” — is the seed from which every property of the circle grows. Every theorem in this chapter is, deep down, just this fact used cleverly: because all radii are equal, triangles you build inside a circle turn out to be congruent, and congruent triangles hand you equal sides and equal angles for free.

The Parts of a Circle

Before we can talk about a circle, we need names for its parts. Everything here happens on a flat sheet of paper — a two-dimensional plane.

A circle is the set of all points on the plane that are the same distance from a fixed point. That fixed point is the centre. The fixed distance from the centre to any point on the circle is the radius.

Mathematicians have a fancy word for “the set of all points that follow a rule”: the locus. So we can also say a circle is the locus of points at a fixed distance from the centre. Same idea, shorter to say.

Now for the lines you can draw. The diagram below names them all, so study it before we go on.

Tap to zoom

Let us pin down each term, using Figure 5.1.

• A chord is a straight line segment joining any two points on the circle. BC is a chord.
• A diameter is a chord that passes through the centre. DE is a diameter. It is the longest chord you can draw (we will prove this later).
• The angle a chord makes at the centre is the angle subtended by the chord at the centre. Chord BC subtends the angle ∠BAC. “Subtend” just means “make an angle across to”. Join the two ends of the chord to the centre, and the angle in between is the subtended angle.

A Circle Is Perfectly Symmetric

A circle has two kinds of symmetry, and both will be useful.

Rotational symmetry. Imagine a spinning wheel on a moving cart. Watch the point of the wheel touching the road. Look away, look back — a point is touching the road again, but you cannot tell if it is the same point. A rotating circle looks identical at every angle of turn. We say the circle has complete rotational symmetry: turn it by any angle about its centre and it looks exactly the same.

Reflection symmetry. Cut out a paper circle. Fold it so the two halves of the boundary land on top of each other. Open it. The crease you made is a line of reflection symmetry — fold along it and the circle matches itself. Does this crease pass through the centre? Yes, always. The crease is a diameter. So every diameter is a line of reflection symmetry, and a circle has infinitely many of them.

This symmetry is not just pretty. It is the quiet reason behind several proofs later: if you rotate the whole circle, a chord just slides to a new position of the same length, carrying its distance-from-centre along with it.

How Many Circles Pass Through Given Points?

Here is a natural question. You are given a couple of points. How many circles can you draw that pass through them?

Through two points — infinitely many

Suppose a circle passes through two points A and B. Its centre, call it O, must satisfy OA = OB (both are radii). So the centre is a point that is the same distance from A and from B.

Which points are equally far from A and B? This is something you met earlier, so here is a quick refresher.

So the centre of any circle through A and B can be any point on the perpendicular bisector of AB. There are infinitely many such points, so there are infinitely many circles. Figure 5.2 shows three of them.

Tap to zoom

The smallest such circle has AB itself as its diameter, with centre at the midpoint M. As the centre slides further out along the bisector, the circles get bigger and bigger, with no largest one.

Through three points — exactly one (the circumcircle)

Now take three points A, B, C. How many circles pass through all three?

First, a warning. If A, B, C lie on one straight line (we say they are collinear), then no circle passes through all three. A straight line can cut a circle in at most two points, never three.

But if the three points are not on a line, something neat happens. There is exactly one circle through them. Let us prove it.

Theorem 1 (Circle through three points). Through any three points that are not on one straight line, there is exactly one circle.

Why is this true? Suppose a circle through A, B, C exists, with centre O.

• Since OA = OB, the centre O lies on the perpendicular bisector of AB.
• Since OA = OC, the centre O also lies on the perpendicular bisector of AC.

So O must be on both bisectors at once — it is their crossing point. Because A, B, C are not collinear, these two bisectors are not parallel, so they cross at exactly one point. That single crossing point is O. Draw the circle with centre O and radius OA, and it passes through all three points. Only one centre is possible, so only one circle is possible. ∎

These three points are also the corners of a triangle. So we have just learned how to draw the circle that passes through all three corners of a triangle. This circle has special names.

• The circle through the three vertices is the circumcircle of the triangle.
• Its centre is the circumcentre.
• We say the circle circumscribes the triangle, and the triangle is inscribed in the circle.

Where the circumcentre lands depends on the shape of the triangle. Figure 5.3 shows the three cases.

Tap to zoom

That last case in Figure 5.3(c) is worth remembering: in a right-angled triangle the circumcentre is the midpoint of the hypotenuse. We will see why near the end of the chapter, when we prove the “angle in a semicircle” result.

Chords and the Angles They Subtend

Tie a thread between two points on a wheel and pull it tight. That thread is a chord. Now turn the wheel. The thread swings to a new spot but keeps its length, so it is a new chord of the same length. The angle it makes at the centre — does that change? It should not, by symmetry. Let us prove it properly.

Equal chords make equal angles at the centre

Theorem 2. Equal chords of a circle subtend equal angles at the centre.

Given: Two chords AB and DE with AB = DE, in a circle with centre C. To show: ∠ACB = ∠DCE.

The plan is to find two congruent triangles whose matching angles are the ones we want. Figure 5.4 shows the set-up.

Tap to zoom

Why is this true? Look at triangles CAB and CDE.

• CA = CB = r (both are radii).
• CD = CE = r (both are radii).
• So CA = CD and CB = CE.
• AB = DE is given.

All three pairs of sides match, so by the SSS congruence rule, △CAB ≅ △CDE. Matching parts of congruent triangles are equal, so ∠ACB = ∠DCE. The equal chords subtend equal angles at the centre. ∎

The converse: equal angles make equal chords

The reverse is also true, and almost as easy.

Theorem 3. If two chords subtend equal angles at the centre, the chords are equal.

Given: ∠ACB = ∠DCE. To show: AB = DE.

This time we still have CA = CB = CD = CE = r. So in triangles ACB and DCE: AC = DC, BC = EC, and the angle between these sides is equal (∠ACB = ∠DCE, given). Two sides and the included angle match, so by the SAS congruence rule, △ACB ≅ △DCE. Therefore AB = DE. ∎

The Centre, the Midpoint, and the Perpendicular

Now we look at what happens when we join the centre to the middle of a chord.

Theorem 4. The line joining the centre of a circle to the midpoint of a chord is perpendicular to that chord.

Given: A circle with centre C, a chord AB, and M the midpoint of AB. To show: CM ⊥ AB (that is, CM meets AB at a right angle).

Figure 5.5 shows the picture.

Tap to zoom

Why is this true? Triangle CAB is isosceles, because CA = CB (both radii). Now compare the two small triangles CMA and CMB:

• CA = CB (radii).
• AM = BM (M is the midpoint).
• CM = CM (it is shared by both).

Three pairs of sides match, so by SSS congruence, △CMA ≅ △CMB. Hence ∠CMA = ∠CMB. But these two angles sit side by side on the straight line AB, so they add up to 180°. Two equal angles adding to 180° must each be 90°. So CM ⊥ AB. ∎

The reverse statement is also true and follows from almost the same picture.

Theorem 5. The perpendicular drawn from the centre of a circle to a chord bisects the chord.

Why: Drop the perpendicular CM from C to chord AB, so ∠CMA = ∠CMB = 90°. In the right triangles CMA and CMB: the hypotenuses CA and CB are equal radii, the side CM is shared, and both have a right angle at M. By the RHS congruence rule the triangles are congruent, so AM = BM. The foot of the perpendicular is the midpoint. ∎

These two facts together are a powerful tool. From the centre, “go to the midpoint” and “drop a perpendicular” lead to the same line. This is the line we measure when we ask “how far is the chord from the centre?”.

Equal Chords Sit at Equal Distances

We have a way to measure how far a chord is from the centre: the length of the perpendicular from the centre to the chord. Now a guess — if two chords have the same length, are they the same distance from the centre? Spin the circle in your mind: a chord just slides to a new place of the same length, dragging its distance along. So they should be equidistant. But sliding pictures are not a proof. Let us prove it.

Theorem 6. Chords of equal length are at equal distances from the centre.

Given: A circle with centre C; chords AB and FG with AB = FG; E and H are the midpoints of AB and FG. To show: CE = CH (the two perpendicular distances are equal).

Figure 5.6 shows the situation.

Tap to zoom

Why is this true? Here is the quickest of two routes. Compare the big triangles CAB and CFG:

• CA = CF (both radii).
• CB = CG (both radii).
• AB = FG (given).

By SSS congruence, △CAB ≅ △CFG. In congruent triangles, the matching heights (altitudes) are also equal. CE is the height of triangle CAB onto AB, and CH is the height of triangle CFG onto FG. So CE = CH. The equal chords are the same distance from the centre. ∎

The converse holds too, and it is what you confirm by Pythagoras: if r² = d² + (half-chord)² and the distances d are equal with the same radius, then the half-chords are equal, so the chords are equal.

Theorem 7. Chords that are equidistant from the centre have equal length.

Which Chord Is Closer — the Long One or the Short One?

Draw a few chords of different lengths and measure each one’s distance from the centre. A clear pattern shows up: the longer the chord, the closer it is to the centre. Let us understand why.

Theorem 8. If two chords of a circle are unequal, the longer one is nearer to the centre.

Given: Chords AB and DE with AB > DE. Drop perpendiculars CF (to AB) and CG (to DE). To show: CF < CG (the longer chord’s distance is smaller).

Figure 5.7 shows the idea.

Tap to zoom

Why is this true? The perpendiculars hit the chords at their midpoints F and G. So in the two right triangles we have, by Pythagoras:

• For chord AB: r² = CF² + AF², where AF is half of AB.
• For chord DE: r² = CG² + GD², where GD is half of DE.

Both equal r², so CF² + AF² = CG² + GD².

Now AB > DE, so half of AB is bigger than half of DE: AF > GD. That means AF² > GD². To keep both sides equal, CF² must be smaller than CG². So CF < CG. The longer chord is nearer the centre. ∎

Two nice end-points of this rule:

• The chord containing the centre is the diameter. Its distance from the centre is 0 — it cannot get any closer. So the diameter is the longest chord. (This proves the claim we made at the start.)
• Push a chord further and further from the centre and it shrinks. When its distance equals the radius, the chord has shrunk to a single point — length 0.

This worked example also shows the handy general formula. If a chord is at distance d from the centre of a circle of radius r, its length is 2 × √(r² − d²).

Arcs: the Curved Pieces of a Circle

An arc is a piece of the circle itself — a connected portion of the curved boundary. Pick two points on a circle. They split the boundary into two arcs: a shorter one and a longer one. The shorter is the minor arc; the longer is the major arc.

Tap to zoom

The angle an arc subtends at the centre is the angle ∠AOB you sweep through as you travel from A to B along that arc. For the minor arc, this is the small angle (less than 180°). For the major arc, you sweep the other way, through the bigger angle (more than 180°). A quick test: if the central angle is less than 180° the arc is minor; if more than 180°, it is major.

The Most Beautiful Fact: Angle at the Centre vs Angle on the Circle

Here is the result that makes circles truly special, the one that unlocks all the rest.

Take an arc. Now take any point on the circle outside that arc, and look at the angle the arc’s two ends make there. Amazingly, this angle is the same no matter which such point you choose — and it is always exactly half the angle the arc makes at the centre.

Theorem 9 (Central Angle Theorem). The angle an arc subtends at the centre is double the angle it subtends at any point on the circle outside the arc.

Given: An arc AFB. C is the centre, so ∠ACB is the angle at the centre. D is any point on the circle outside arc AFB. To show: ∠ACB = 2 × ∠ADB.

Figure 5.9 shows the main case.

Tap to zoom

Why is this true? Join D to the centre C, and extend that line to meet the circle again at E. The trick is that the line CD splits both angles into two halves we can handle separately, using isosceles triangles.

Look at triangle DCB. Here CB = CD (both radii), so it is isosceles, which means ∠CBD = ∠CDB. Now ∠BCE is the exterior angle of triangle BCD at C.

So, using the exterior-angle rule on triangle BCD: ∠BCE = ∠CBD + ∠CDB = 2 × ∠CDB (since the two base angles are equal).

Do the same with triangle ACD: CA = CD (radii), so it is isosceles, and ∠ACE is its exterior angle. Hence: ∠ACE = ∠CAD + ∠CDA = 2 × ∠CDA.

Now just add the two halves. The full central angle is ∠ACB = ∠BCE + ∠ECA = 2∠CDB + 2∠CDA = 2(∠CDB + ∠CDA) = 2 × ∠BDA. So:

∠ACB = 2 × ∠ADB. ∎

(When D sits so that the line CD meets the circle outside the arc instead of inside it, the same isosceles-triangle idea works — you just subtract the two pieces instead of adding them, and the answer comes out the same.)

What this gives us: angles in the same arc segment are equal

The angle at the centre is one fixed number. Every point on the circle outside the arc sees the arc at half of that fixed number. So all those points see the arc at the same angle. If you stand at any point on the major arc and look at chord AB, the angle is the same wherever you stand. This equal-angle property is unique to the circle — no other shape does it.

The angle in a semicircle is 90°

A beautiful special case falls right out.

Corollary. The angle in a semicircle is a right angle. (A corollary is a fact that drops out immediately from a result you have already proved.)

If AB is a diameter, take any point D on the circle and look at ∠ADB. Figure 5.10 shows it.

Tap to zoom

Why: The diameter AB passes through the centre C, so the angle the arc (the half not containing D) subtends at the centre is a straight angle, ∠ACB = 180°. By Theorem 9, the angle at D is half of that: ∠ADB = ½ × 180° = 90°. This holds for every point D on the circle. ∎

This also explains Figure 5.3(c): in a right triangle the hypotenuse must be a diameter of the circumcircle, so the circumcentre is its midpoint.

When Do Four Points Lie on One Circle?

We saw three non-collinear points always lie on one circle. But four points usually do not. So when do four points lie on a single circle? Points that share one circle are called concyclic.

Theorem 10. If a segment AB subtends equal angles at two points C and D on the same side of AB, then A, B, C, D all lie on one circle.

Given: ∠ACB = ∠ADB, with C and D on the same side of AB (and neither on the line AB). To show: A, B, C, D are concyclic.

Why is this true? This is a clever proof by contradiction. The three points A, B, C are not collinear, so by Theorem 1 there is a unique circle through them. We claim D is on it too. Suppose not. Then D is either inside or outside this circle.

Join A to D. If D is outside, the line AD crosses the circle at some point E between A and D. Now C and E are both on the same arc cut off by chord AB, so by the “angles in the same arc are equal” result, ∠AEB = ∠ACB. But ∠AEB is an exterior angle of triangle BED, so it must be bigger than the far interior angle ∠ADB. Putting it together: ∠ACB = ∠AEB > ∠ADB. Yet we were given ∠ACB = ∠ADB. So ∠ACB would be bigger than itself — impossible.

The case “D inside the circle” leads to the same kind of impossibility. Since D cannot be inside and cannot be outside, D must lie on the circle. So all four points are concyclic. ∎

When the four corners of a quadrilateral all lie on a circle, we call it a cyclic quadrilateral. It has a lovely property.

Opposite angles of a cyclic quadrilateral

Theorem 11. In a cyclic quadrilateral, each pair of opposite angles adds up to 180°.

Given: A, B, C, D lie on a circle with centre O, forming cyclic quadrilateral ABCD. To show: ∠A + ∠C = 180° (and likewise ∠B + ∠D = 180°).

Figure 5.11 shows a cyclic quadrilateral with its two opposite angles marked.

Tap to zoom

Why is this true? We use the central angle theorem. The angle ∠A (that is, ∠BAD) is the angle that arc BCD subtends at the point A on the circle. By Theorem 9 it is half the angle that arc BCD subtends at the centre. As you sweep from OB round to OD passing C, that is the reflex angle BOD (the bigger one). So:

∠BAD = ½ × (reflex ∠BOD).

Similarly, ∠C (that is, ∠BCD) is half the angle the other arc, BAD, subtends at the centre — the ordinary (non-reflex) ∠BOD:

∠BCD = ½ × (∠BOD).

Now add them. The reflex ∠BOD and the ordinary ∠BOD together make one full turn around the centre, which is 360°. So:

∠BAD + ∠BCD = ½ × (reflex ∠BOD + ∠BOD) = ½ × 360° = 180°. ∎

The reverse is also true (we will not write out the full proof, but it follows the same contradiction idea as Theorem 10):

Theorem 12. If one pair of opposite angles of a quadrilateral adds up to 180°, then the quadrilateral is cyclic — all four corners lie on a circle.

Common Mistakes

Quick Check

Practice Problems

Easy

Medium

Challenge

Summary

You can now explain:

• What a circle, centre, radius, chord, diameter and arc are, and what it means for a chord or arc to subtend an angle at the centre.
• Why a circle has complete rotational symmetry and why every diameter is a line of reflection symmetry.
• Why infinitely many circles pass through two points (centres on the perpendicular bisector), but exactly one passes through three non-collinear points (the circumcircle), and where its centre lands for acute, obtuse and right triangles.
• The chord-and-angle theorems: equal chords subtend equal angles at the centre (and the converse), the line from the centre to a chord’s midpoint is perpendicular to it (and the converse), and equal chords are equidistant from the centre (and the converse).
• Why the longer of two chords is always nearer the centre, using length = 2√(r² − d²).
• The central angle theorem (angle at centre = twice the angle on the circle), and its consequences: angles in the same arc are equal, the angle in a semicircle is 90°, and the opposite angles of a cyclic quadrilateral add to 180°.
• How to use these results to find unknown chord lengths, distances and angles in exam-style problems.

What’s Next

You have measured the angles and distances hidden inside circles. Next you will measure their size — how much boundary a shape has and how much flat space it covers. In Chapter 6 — Measuring Space: Perimeter and Area you will work with the perimeter and area of triangles, quadrilaterals and circles, and meet Heron’s formula for the area of any triangle. The Pythagoras theorem you used so often here will come up again, so keep it handy.