I'm Up and Down, and Round and Round

Every raindrop that lands on still water sends out a ripple that is, for an instant, a perfect circle. Every wheel that has ever turned traces one. Cave paintings at Gudahandi in Odisha already place circles beside triangles and squares — a shape humans noticed in nature long before geometry gave it a name. The full moon, the sun during an eclipse, the cross-section of a plant stem: circles are everywhere, and this chapter turns that everyday shape into a precise mathematical object. The whole chapter grows out of one sentence: a circle is the set of all points in a plane that are equidistant from a fixed point, its centre. That distance is the radius, and a segment joining two points on the circle is a chord. From this single definition you will build an entire toolkit — why a circle looks the same after any rotation and after reflection in any diameter, why exactly one circle passes through three non-collinear points (giving every triangle its own circumcircle), and how that circle's centre can sit inside, outside, or exactly on the triangle depending on its angles. The heart of the chapter is a chain of paired results about chords: equal chords subtend equal angles at the centre, equal chords sit at equal distances from the centre, and the longer of two chords always sits closer to the centre. A separate strand studies arcs and the angles they subtend — culminating in the striking fact that an arc's angle at the centre is always exactly double its angle anywhere else on the circle, which is why every angle drawn in a semicircle is a right angle. The chapter closes by asking when four points can lie on one circle at all, leading to the elegant rule that governs every cyclic quadrilateral: its opposite angles always add up to a straight angle.

Key takeaways

A circle is the locus of points in a plane equidistant from a fixed centre; that fixed distance is the radius, a segment joining two points on the circle is a chord, and a chord through the centre is a diameter.
A circle has complete rotational symmetry about its centre and reflection symmetry across every diameter.
Infinitely many circles pass through two given points; their centres lie on the perpendicular bisector of the segment joining the points.
Theorem 1: exactly one circle — the circumcircle — passes through three non-collinear points; its centre, the circumcentre, is where two perpendicular bisectors of the triangle's sides meet.
Theorems 2-3: equal chords subtend equal angles at the centre, and conversely.
Theorems 4-5: the segment from the centre to a chord's midpoint is perpendicular to it, and conversely, the perpendicular from the centre bisects the chord.
Theorems 6-8: equal chords are equidistant from the centre (and conversely); of two unequal chords, the longer is closer to the centre; a chord of length $\ell$ at distance $d$ satisfies $\ell = 2\sqrt{r^2 - d^2}$ .
Theorem 9: the angle an arc subtends at the centre is double the angle it subtends anywhere on the remaining circle; a corollary is that the angle in a semicircle is always $90^\circ$ .
Theorems 10-12: equal angles on the same side of a segment make four points concyclic; opposite angles of a cyclic quadrilateral sum to $180^\circ$ , and conversely.
Key idea: every property of a circle in this chapter — symmetry, chords, arcs, cyclic quadrilaterals — traces back to one fact, that all points on the circle sit at the same fixed distance from a single centre.

The core insight

Because every point of a circle lies at the same fixed distance from its centre, congruent triangles built from radii and chords force a chain of paired results — equal chords subtend equal angles and sit equidistant from the centre, arcs subtend angles at the centre exactly double those at the circumference, and quadrilaterals are cyclic precisely when opposite angles sum to $180^\circ$ .

What this chapter covers

The chapter opens with the definition of a circle as a locus, names its parts — centre, radius, chord, diameter — and shows its two symmetries: rotational (about the centre, through any angle) and reflective (across any diameter). It then asks how many circles pass through a given set of points: infinitely many through two, but exactly one, the circumcircle, through three non-collinear points, the source of every triangle's circumcentre. The largest block studies chords: equal chords and the angles they subtend at the centre, how the perpendicular from the centre bisects a chord, and the exact formula linking radius, distance and chord length. It then turns to arcs — minor, major, and the angle each subtends at the centre and elsewhere on the circle — reaching the chapter's central theorem, that the angle at the centre is double the angle at the circumference, with the right angle in a semicircle as its most famous case. It closes with concyclic points and the signature rule of a cyclic quadrilateral: opposite angles always add to $180^\circ$ .

The core ideas

Circle, chord, diameter — the basic vocabulary

A circle is the set of all points on a plane that are equidistant from a given point on that plane; this set of points is also called the locus of points satisfying the given condition. The fixed point is the centre, and the common distance from the centre to any point on the circle is the radius. If $A$ is the centre and $B$ is any point on the circle, then every point at the same distance $AB$ from $A$ also lies on the circle.

\text{Circle with centre } A, \text{ radius } r: \quad \{P : AP = r\}. \qquad (1)

where: $P$ ranges over every point of the plane, and only those points at exact distance $r$ from $A$ belong to the circle. If $B$ and $C$ are two points on the circle, the line segment $BC$ is a chord, and the angle it makes at the centre is $\angle BAC$ . A chord that passes through the centre is called a diameter; a diameter is automatically the longest possible chord, since the centre itself lies on it and the two radii $AB$ and $AC$ line up into a straight segment of length $2r$ . Fold a paper circle so its boundary overlaps itself; the crease always passes through the centre, so it is a diameter — a line of reflection symmetry. Every diameter gives such a crease, so a circle has infinitely many. Spin the circle about its centre by any angle and it looks unchanged — complete rotational symmetry, unlike a square (repeats every $90^\circ$ ) or a regular hexagon (every $60^\circ$ ).

Circles through two points, and the special circle through three

Given two points $A$ and $B$ , how many circles pass through both? Any such circle has a centre $O$ with $OA = OB$ , so $O$ is equidistant from $A$ and $B$ — meaning $O$ lies on the perpendicular bisector of $AB$ . Conversely every point on that bisector is equidistant from $A$ and $B$ , so it is exactly the set of possible centres. A line holds infinitely many points, so infinitely many circles pass through any two given points, one for each choice of centre on the bisector. Three non-collinear points $A$ , $B$ , $C$ pin this down completely.

\textbf{Theorem 1.} \ \text{A unique circle passes through three non-collinear points.} \qquad (2)

where: "non-collinear" rules out three points on one line, through which no circle can pass at all. The full reasoning is in the first derivation below; the resulting circle is the circumcircle of $\triangle ABC$ , and its centre $O$ is the circumcentre, the intersection of the perpendicular bisectors of the three sides. Its location depends on the triangle's shape: inside for an acute triangle, outside for an obtuse triangle, and exactly at the midpoint of the hypotenuse for a right triangle.

Figure

Three triangles acute obtuse and right angled each with its circumcircle and circumcentre marked showing the centre moves inside outside or onto the hypotenuse

Equal chords, equal angles at the centre

Tie a thread across two points on a wheel and pull it tight — that thread is a chord, subtending an angle at the centre. Rotate the wheel: the thread traces a new chord of the same length, and it is natural to guess it subtends the same angle. This pairing is proved by comparing triangles built from two radii and a chord.

\textbf{Theorem 2. } AB = DE \implies \angle ACB = \angle DCE. \qquad \textbf{Theorem 3 (converse). } \angle ACB = \angle DCE \implies AB = DE. \qquad (3)

where: $C$ is the centre and $AB$ , $DE$ are chords. Theorem 2 follows because $CA = CB = CD = CE = r$ (all radii), so with $AB = DE$ given, $\triangle CAB \cong \triangle CDE$ by SSS and the included angles match. Theorem 3 reverses it: with $CA = CD$ , $CB = CE$ and $\angle ACB = \angle DCE$ given, $\triangle CAB \cong \triangle CDE$ by SAS, so $AB = DE$ .

Perpendiculars from the centre: bisecting chords and measuring their distance

Draw any chord $AB$ of a circle with centre $C$ , and let $M$ be its midpoint. Triangle $CAB$ is isosceles because $CA = CB = r$ , so $\angle CAB = \angle CBA$ . Since $AM = BM$ and $CM$ is shared, $\triangle CMA \cong \triangle CMB$ by SAS, so $\angle CMA = \angle CMB$ ; being angles on a straight line, each is $90^\circ$ .

\textbf{Theorem 4. } CM \perp AB. \qquad \textbf{Theorem 5 (converse). } CM \perp AB \implies AM = BM. \qquad (4)

where: Theorem 5 reads the same congruent triangles the other way: given $\angle CMA = \angle CMB = 90^\circ$ , shared $CM$ and $CA = CB$ , $\triangle CMA \cong \triangle CMB$ by RHS, so $AM = BM$ . The segment from the centre to a chord's midpoint and the perpendicular from the centre to the chord are one and the same line — the standard way to locate a circle's centre from any chord. This gives the distance of a chord from the centre: the length $CM$ . In right triangle $CMA$ , the Baudhāyana-Pythagoras Theorem gives $r^2 = d^2 + (\ell/2)^2$ where $d = CM$ is the distance and $\ell = AB$ is the chord length, so

\ell = 2\sqrt{r^2 - d^2}. \qquad (5)

where: $r$ is the radius, $d$ the perpendicular distance from the centre, $\ell$ the chord length. This one formula packages Theorems 6-8: since $\ell$ falls as $d$ rises, equal chords share the same $d$ (6, 7 converse), and a longer chord forces a smaller $d$ , so the longer of two chords is always closer to the centre (8). At $d = 0$ , $\ell = 2r$ , the diameter, the longest possible chord.

Arcs and the angle they subtend

An arc is a connected portion of the circle cut off by two points on it. Two points $A$ , $B$ split the circle into two arcs; the larger piece is the major arc and the smaller the minor arc. The angle subtended by an arc at the centre is the angle swept moving one radius to the other along that arc: the minor arc $AXB$ subtends $\angle AOB$ the short way, and the major arc $AYB$ subtends it the long way round — the reflex angle, $360^\circ - \angle AOB$ .

The angle-at-the-centre theorem and angles in the same segment

Fix an arc $AB$ and a point $D$ on the circle outside it; $\angle ADB$ is the angle the arc subtends at $D$ . Remarkably this angle does not depend on where $D$ sits, so long as $D$ stays outside the chosen arc.

\textbf{Theorem 9. } \angle AOB = 2\, \angle ADB, \quad \text{for any } D \text{ on the circle outside arc } AB. \qquad (6)

where: $O$ is the centre. The full argument, via the exterior-angle property of isosceles triangles, is in the second derivation below. Two consequences follow at once. A diameter subtends a straight angle, $180^\circ$ , at the centre, so it subtends $90^\circ$ everywhere on the circle — the angle in a semicircle is a right angle. And since $\angle ADB$ is always exactly half the fixed $\angle AOB$ , every such $D$ on the same side of $AB$ gives the same angle: angles in the same segment of a circle are equal.

Figure

A circle with centre O an arc AB subtending angle AOB at the centre and two points D and E on the major arc each showing angle ADB and angle AEB equal to half of angle AOB with a diameter case showing a right angle

Concyclic points and the cyclic quadrilateral

Points on the same circle are called concyclic. Theorem 9 gives a test: if a segment $AB$ subtends equal angles at two other points $C$ , $D$ on the same side of $AB$ , then $A$ , $B$ , $C$ , $D$ all lie on one circle. When a quadrilateral's four vertices are concyclic it is called cyclic, and it inherits a striking angle rule.

\textbf{Theorem 11. } \angle BAD + \angle BCD = 180^\circ, \quad \text{for a cyclic quadrilateral } ABCD. \qquad (7)

where: the reasoning uses Theorem 9 twice: $\angle BAD$ is half the reflex angle $\angle BOD$ (as $A$ sits outside arc $BCD$ ), and $\angle BCD$ is half the ordinary $\angle BOD$ (as $C$ sits outside arc $BAD$ ). Adding the two halves accounts for the full angle at the centre, $360^\circ$ , so $\angle BAD + \angle BCD = 180^\circ$ ; the same argument gives $\angle ABC + \angle ADC = 180^\circ$ . Theorem 12, the converse, says a quadrilateral whose opposite angles sum to $180^\circ$ must be concyclic — proved by contradiction, placing a fourth point off the circle through the other three and showing it forces an angle to exceed itself.

Derivation AA unique circle passes through three non-collinear points

The goal is to show that for any three points $A$ , $B$ , $C$ not on one straight line, exactly one circle passes through all three — not zero, and not more than one — using nothing beyond the definition of a circle and the fact that two intersecting lines meet at a single point. This is what gives every triangle its own circumcircle, and pinning down existence and uniqueness together is the whole point.

Step 1. Suppose a circle does pass through $A$ , $B$ , $C$ ; call its unknown centre $O$ . Since $O$ is the centre, $OA = OB$ , so $O$ is equidistant from $A$ and $B$ , and any point equidistant from two given points lies on their perpendicular bisector. So $O$ lies on the perpendicular bisector of $AB$ .

Step 2. The same reasoning applied to the pair $A$ , $C$ gives $OA = OC$ , so $O$ must also lie on the perpendicular bisector of $AC$ . Any working centre has to sit on both bisectors at once.

Step 3. Since $A$ , $B$ , $C$ are not collinear, the perpendicular bisectors of $AB$ and $AC$ are neither identical nor parallel, so they meet in exactly one point. Call it $O$ ; it is forced to be the centre by Steps 1 and 2, and there is only one such point.

Step 4. Existence: draw the circle centred at $O$ with radius $OA$ . Because $O$ is on the perpendicular bisector of $AB$ , $OA = OB$ , so $B$ lies on this circle; because $O$ is on the bisector of $AC$ , $OA = OC$ , so $C$ lies on it too. This one circle passes through all three points.

Step 5. Uniqueness follows from Step 3: the centre of any circle through $A$ , $B$ , $C$ must be that single intersection point, so no second, different circle through all three can exist. $O$ is the circumcentre of $\triangle ABC$ , and the circle is its circumcircle.

Go deeper · Derivation B · 1 minThe angle an arc subtends at the centre is double the angle it subtends at the circumference

Derivation BThe angle an arc subtends at the centre is double the angle it subtends at the circumference

The goal is to prove that for an arc $AFB$ of a circle with centre $C$ , and any point $D$ on the circle outside that arc, the angle $\angle ACB$ at the centre is exactly twice the angle $\angle ADB$ at $D$ — using only the exterior-angle property of a triangle (an exterior angle equals the sum of the two non-adjacent interior angles), applied twice to isosceles triangles built from radii. We treat the case where line $DC$ , extended, meets the circle again at a point $E$ on arc $AFB$ itself; the case where $E$ falls outside that arc uses the same two steps with a subtraction in place of an addition.

Step 1. Join $D$ to $C$ and extend to meet the circle again at $E$ , on arc $AFB$ . In $\triangle DCB$ , $CB = CD$ (both radii), so the triangle is isosceles with $\angle CBD = \angle CDB$ .

Step 2. $\angle BCE$ is an exterior angle of $\triangle DCB$ at $C$ , lying on the straight line $DCE$ . By the exterior angle property, $\angle BCE = \angle CBD + \angle CDB = 2\,\angle BDC$ .

Step 3. Repeat on the other side with $\triangle DCA$ . Since $CA = CD$ , $\angle CAD = \angle CDA$ , and $\angle ACE$ , an exterior angle of $\triangle DCA$ , equals $\angle CAD + \angle CDA = 2\,\angle CDA$ .

Step 4. Since $\angle BCA = \angle BCE + \angle ECA$ and $\angle BDA = \angle BDC + \angle CDA$ (as $E$ lies between rays $DA$ and $DB$ ), adding Steps 2 and 3 gives $\angle BCA = 2(\angle BDC + \angle CDA) = 2\,\angle BDA$ — exactly the claim.

Step 5. When $E$ falls outside arc $AFB$ instead, the same two isosceles triangles give $\angle ACE = 2\,\angle ADC$ and $\angle BCE = 2\,\angle BDC$ , but now $\angle ACB = \angle ACE - \angle BCE$ and $\angle ADB = \angle ADC - \angle BDC$ , since $E$ no longer lies between the two rays. Subtracting instead of adding gives the same conclusion, $\angle ACB = 2\,\angle ADB$ , so the doubling relation holds in every configuration.

Worked examples

Board level - chord length from radius and distance from the centre

A chord of a circle is $5\ \text{cm}$ away from the centre. If the radius of the circle is $13\ \text{cm}$ , find the length of the chord.

Step 1. The perpendicular from the centre to the chord bisects it (Theorem 5), creating a right triangle whose hypotenuse is the radius, one leg is the distance $d = 5\ \text{cm}$ , and the other leg is half the chord.

Step 2. By the Baudhāyana-Pythagoras Theorem, $(\ell/2)^2 = r^2 - d^2 = 13^2 - 5^2 = 169 - 25 = 144$ , so $\ell/2 = 12\ \text{cm}$ .

Step 3. Double it for the full chord: $\ell = 2 \times 12 = 24\ \text{cm}$ . Answer: the chord is $24\ \text{cm}$ long.

Board level - opposite angles of a cyclic quadrilateral

$ABCD$ is a cyclic quadrilateral. If $\angle A = 75^\circ$ , find $\angle C$ . If $\angle B = 110^\circ$ , find $\angle D$ .

Step 1. By Theorem 11, opposite angles of a cyclic quadrilateral sum to $180^\circ$ , so $\angle A + \angle C = 180^\circ$ .

Step 2. Substituting, $\angle C = 180^\circ - 75^\circ = 105^\circ$ . Applying the same rule to the other pair, $\angle B + \angle D = 180^\circ$ , so $\angle D = 180^\circ - 110^\circ = 70^\circ$ . Answer: $\angle C = 105^\circ$ and $\angle D = 70^\circ$ .

Higher-order thinking (HOTS / olympiad-style) - radius from two parallel chords on the same side of the centre

Two parallel chords of a circle have lengths $10\ \text{cm}$ and $24\ \text{cm}$ and lie on the same side of the centre. The distance between the two chords is $7\ \text{cm}$ . Find the radius of the circle.

Step 1. Let $O$ be the centre, with perpendicular distances $d_1$ to the $10\ \text{cm}$ chord and $d_2$ to the $24\ \text{cm}$ chord. Since the shorter chord is farther from the centre (Theorem 8), $d_1 > d_2$ , and because both chords lie on the same side of $O$ , the gap between them is the difference of the distances: $d_1 - d_2 = 7$ .

Step 2. Apply formula (5) to each chord, using half-lengths $5$ and $12$ : $r^2 = d_1^2 + 5^2$ and $r^2 = d_2^2 + 12^2$ . Substitute $d_1 = d_2 + 7$ into the first equation and set it equal to the second: $(d_2+7)^2 + 25 = d_2^2 + 144$ .

Step 3. Expand and simplify: $d_2^2 + 14d_2 + 49 + 25 = d_2^2 + 144$ , so $14 d_2 = 144 - 74 = 70$ , giving $d_2 = 5\ \text{cm}$ and hence $d_1 = 12\ \text{cm}$ .

Step 4. Substitute back: $r^2 = d_2^2 + 12^2 = 25 + 144 = 169$ , so $r = 13\ \text{cm}$ . As a check, the other equation gives $r^2 = d_1^2 + 5^2 = 144 + 25 = 169$ too. Answer: the radius is $13\ \text{cm}$ . Timing note: about three minutes. The whole problem is two applications of formula (5) tied together by one linear relation between the two unknown distances — the trap is deciding the sign of that relation correctly, which is why Theorem 8 (longer chord, smaller distance) must be applied first to fix which distance is bigger, before writing $d_1 - d_2 = 7$ rather than $d_2 - d_1 = 7$ . Setting up the two Pythagorean equations correctly and eliminating $r^2$ between them turns the geometry into routine algebra.

Common traps

Mixing up the direction of the angle-doubling rule is the most common slip: the angle at the centre is double the angle at the circumference, never the reverse — read the figure carefully before applying Theorem 9. Forgetting that Theorems 2-3, 4-5 and 6-7 are genuine "if and only if" pairs costs marks: proving one direction (equal chords give equal angles) does not prove the converse, and exam questions often ask for it specifically. Using $\ell = 2\sqrt{r^2 - d^2}$ with $d$ measured as anything other than the perpendicular distance from the centre gives a wrong answer — only the perpendicular forms a right angle with the chord. Assuming any four points lie on a circle, or that any quadrilateral is cyclic — concyclicity is a special condition (equal angles on one side, or opposite angles summing to $180^\circ$ ) that must be checked, not assumed. Confusing the angle an arc subtends at the centre with the acute angle between the radii when the true angle is reflex — for a major arc the swept angle exceeds $180^\circ$ , and missing this gives the wrong half in a cyclic-quadrilateral calculation. Treating the circumcentre as always inside the triangle: it is inside only for an acute triangle, at the hypotenuse's midpoint for a right triangle, and genuinely outside for an obtuse one.

Common trapThe angle at the centre is always double the angle at the circumference, never equal to it

Students otherwise fluent with congruent triangles often still write $\angle ADB = \angle AOB$ instead of $\angle ADB = \tfrac{1}{2}\angle AOB$ , especially when a figure is unfamiliar. The doubling is not a rounding convenience — it comes directly from two exterior-angle arguments stacked together (see the second derivation), and it is why the angle in a semicircle is $90^\circ$ rather than $180^\circ$ , and why opposite angles of a cyclic quadrilateral sum to $180^\circ$ rather than $360^\circ$ . Whenever a problem mentions an angle "at the centre" and an angle "on the circle" for the same arc, write the factor of two explicitly before any further algebra.

You should be able to…

state the definition of a circle as a locus, and name its centre, radius, chord and diameter
explain why a circle has rotational symmetry about its centre, and reflection symmetry across every diameter
explain why infinitely many circles pass through two points, with centres on their perpendicular bisector
state and apply Theorem 1, constructing a triangle's circumcircle, and predict whether the circumcentre lies inside, outside, or on the triangle
prove and apply the paired results linking equal chords to equal angles at the centre, and to equal distance from the centre
use $\ell = 2\sqrt{r^2 - d^2}$ to find a chord's length, its distance from the centre, or the radius, given the other two
distinguish a minor arc from a major arc, and state the angle each subtends at the centre
state and apply Theorem 9, that the angle at the centre is double the angle at the circumference, including the semicircle corollary
test whether four points are concyclic using equal angles subtended on one side of a segment
state and apply the cyclic quadrilateral rule, that opposite angles sum to $180^\circ$ , in both directions

Ask the tutor: Ask the tutor to walk through constructing the circumcircle of your own triangle, or to check your working on a chord-and-distance problem step by step.

Why does the perpendicular bisector of a chord always pass through the centre?

Because the centre is, by definition, equidistant from every point on the circle, and in particular from the two endpoints of any chord. Any point equidistant from two given points must lie on their perpendicular bisector — that is exactly what a perpendicular bisector is. Since the centre is equidistant from both endpoints of the chord, it has no choice but to lie on that chord's perpendicular bisector. This single fact is also the practical way to find the centre of an unmarked circular object: draw any two chords, construct their perpendicular bisectors, and the point where the two bisectors cross is the centre, since it is the one point common to both.

Why is the circumcentre sometimes outside the triangle?

The circumcentre is simply the intersection of the perpendicular bisectors of the triangle's sides, and that intersection point does not have to sit inside the triangle itself. For an acute triangle all three angles are less than $90^\circ$ and the bisectors cross inside. As one angle grows past $90^\circ$ in an obtuse triangle, the intersection point is pushed outside the triangle, across the side opposite the obtuse angle. Exactly at $90^\circ$ , the right-angle case, the circumcentre lands precisely on the midpoint of the hypotenuse — a useful shortcut, since it means the hypotenuse of a right triangle is always a diameter of its circumcircle.

What exactly is the difference between the angle at the centre and the angle at the circumference for the same arc?

Both angles are measured at the two endpoints of the same chord, but from two different points: one from the centre, and one from any point on the circle that is outside the arc in question. Theorem 9 says the centre angle is always exactly twice the circumference angle, for every choice of the point on the circle. This is why the term "the" angle at the circumference makes sense even though there are infinitely many possible points to choose — they all give the identical angle, since they are all half of the one fixed angle at the centre. Only when the chosen point moves to the other arc does the angle change, becoming the supplement of the first, which is exactly the cyclic-quadrilateral rule.

How do I remember which chord is closer to the centre when two chords have different lengths?

Think of the diameter as the benchmark: it is the longest possible chord, and it passes exactly through the centre, so its distance from the centre is zero. As a chord shrinks away from that maximum length, it has to move farther from the centre, until in the extreme case a "chord" shrinks to a single point on the circle, whose distance from the centre equals the full radius. So length and distance move in opposite directions — longer chord, smaller distance; shorter chord, larger distance. The formula $\ell = 2\sqrt{r^2 - d^2}$ makes this precise: as $d$ increases, the quantity under the square root shrinks, so $\ell$ must shrink too.