Circles — centre, radius, diameter, chord, arc, sector, segment: Question bank

This page is a stress-test for your understanding, not your arithmetic. Every item below hides a common misconception or a boundary case. Read the prompt, commit to an answer out loud, then reveal. If your reasoning differs from the answer's reasoning — even if the yes/no matches — that's a gap worth closing.

Prerequisite ideas live in the parent note Circles — parts of a circle. Deeper machinery is linked where it helps: Angle Properties in Circles, Circle Theorems, Radian Measure, Sectors and Pizza Problem, Circumference and Area of Circles, Pythagorean Theorem.

Definition The words and symbols we lean on

Before the traps, pin down the vocabulary so the reveals never surprise you:

Centre $O$ : the fixed anchor point. Every edge point is the same distance from it.
Radius $r$ : the centre-to-edge distance, $r = O P$ for any edge point $P$ .
Diameter $d$ : a chord that runs through the centre, so $d = 2 r$ . It is the longest chord.
Chord: a straight segment joining two edge points (does not have to pass through $O$ ).
Central angle $θ$ : the angle at $O$ between two radii, measured in radians (an angle of $1$ radian cuts an arc exactly $r$ long — this is why $s = r θ$ ; see Radian Measure).
Arc: a curved piece of the edge; the minor arc is the shorter piece, the major arc the longer.
Sector: the pie-slice region between two radii and their arc.
Segment: the region between a chord and its arc (a sector with the central triangle removed).

To make these tangible, here is the whole cast in one picture — refer back to it whenever a term appears below.

Figure — Circles — centre, radius, diameter, chord, arc, sector, segment

The two proofs that trip people up (chord bisection, and arc-versus-chord in the small-angle limit) each get their own figure at the point they are discussed.

True or false — justify

Every statement is either true or false. The reveal gives the reason, which is the part that actually matters.

A diameter is the longest chord you can draw in a circle.

True. Any chord's length is limited by how far apart two edge points can be; that maximum happens when the chord passes through

O

, giving length

2 r

. Skip the centre and both endpoints crowd toward the same side, shortening the span.

Every chord is a diameter.

False. A chord only needs its two endpoints on the circle. Only the special chords that also pass through the centre are diameters; all others are strictly shorter than

2 r

The radius and the diameter are two names for the same length.

False. The radius

r

is centre-to-edge; the diameter

d = 2 r

is edge-through-centre-to-edge. They differ by a factor of two, so confusing them halves or doubles every answer.

Arc length

s = r θ

works for any angle

θ

you plug in.

False. It works only when

θ

is measured in radians. Feed it

60

(degrees) instead of

\frac{π}{3}

and you overstate the arc by a factor of

\frac{180}{π} \approx 57

. See Radian Measure.

The perpendicular from the centre to a chord always cuts that chord exactly in half.

True. Call the chord

A B

and let

M

be the foot of the perpendicular dropped from

O

(see the figure in the Why section). The two triangles

O M A

and

O M B

share the side

O M

, both have a right angle at

M

, and both have hypotenuse

O A = O B = r

. By RHS congruence

A M = M B

— this is a standard circle theorem.

A sector and a segment cut off by the same chord have the same area.

False. The sector is the whole "pizza slice" (two radii + arc); the segment is that slice with the central triangle removed. So

A_{segment} = A_{sector} - A_{triangle}

, always smaller than the sector (for

θ < π

Doubling the radius doubles the area of a circle.

False. Area is

π r^{2}

, so area scales with the square of

r

. Double

r

and the area becomes four times as large, not twice.

If a chord gets longer, its distance from the centre gets larger.

False. It's the opposite. Longer chords sit closer to the centre; the longest chord (the diameter) passes right through

O

at distance

0

. Naming the foot of the perpendicular

M

, the half-chord

A M

and the centre-distance

O M

obey

r^{2} = A M^{2} + O M^{2}

via the Pythagorean Theorem, so a bigger

A M

forces a smaller

O M

The minor arc and major arc between two points always add up to the full circumference.

True. Together the two arcs trace the entire edge exactly once, so their lengths sum to

2 π r

and their angles sum to

2 π

radians.

A semicircular arc is both the minor and the major arc at once.

True (as a degenerate case). When the two points are diametrically opposite, each arc is exactly half the circle (

180°

), so neither is "shorter" — the minor/major distinction collapses.

Spot the error

Each line contains a flawed statement or one flawed step. Name the mistake.

"Arc length of a

90°

arc on radius

4

s = r θ = 4 \times 90 = 360

The angle was left in degrees. It must be converted:

θ = 90 \times \frac{π}{180} = \frac{π}{2}

, giving

s = 4 \cdot \frac{π}{2} = 2 π \approx 6.28

, not

360

"The sector area is

\frac{1}{2} r^{2} θ

, so for

θ

in degrees just plug the degrees in."

The formula

\frac{1}{2} r^{2} θ

demands radians. In degrees you must use

\frac{θ}{360} π r^{2}

instead, or convert first with

\times \frac{π}{180}

"A segment's area is

\frac{1}{2} r^{2} θ + \frac{1}{2} r^{2} sin θ

The sign is wrong. You subtract the triangle from the sector:

\frac{1}{2} r^{2} (θ - sin θ)

. Adding would make the segment larger than the whole slice, which is impossible.

"Since

d = 2 r

, the circumference is

C = π r

Circumference is

C = π d = 2 π r

, not

π r

. Substituting

d = 2 r

into

C = π d

gives

2 π r

— the person forgot the factor of two lives in the diameter. See Circumference and Area of Circles.

"The triangle inside a sector has area

\frac{1}{2} r^{2}

Only if

θ = 90°

where

sin θ = 1

. In general the two radii meet at angle

θ

, so the area is

\frac{1}{2} r^{2} sin θ

— it shrinks toward

0

θ

shrinks.

"A chord of length

2 r

isn't a diameter because it doesn't have to pass through the centre."

The only way two edge points can be

2 r

apart is if the segment joining them passes through

O

; that maximal separation forces it through the centre, so a length-

2 r

chord is a diameter.

"Points inside the circle are on the circle too, since they're near the centre."

The circle is only the edge — the set of points at distance exactly

r

. Points nearer than

r

form the interior (disk), which is a different set.

"The major arc for a central angle

θ

has length

r θ

, same formula as the minor arc."

No —

r θ

is the minor arc. The major arc is the rest of the edge, spanning angle

2 π - θ

, so its length is

r (2 π - θ) = 2 π r - r θ

Why questions

These probe the reason behind a rule. A memorised formula that can't answer "why" isn't understood yet.

Why must

θ

be in radians for

s = r θ

Radians are defined so that an angle equals its arc length divided by

r

; that's exactly the relationship

θ = s / r

rearranged. Degrees carry no such built-in link to length, so they need a conversion first.

Why is the diameter the longest chord and not some slanted chord?

A chord's length is the straight-line distance between two edge points, and no two points on the circle can be farther apart than

2 r

. That extreme is reached precisely when the line runs through the centre.

Why does

A_{segment} = \frac{1}{2} r^{2} (θ - sin θ)

use

sin θ

The subtracted triangle has two sides

r

meeting at angle

θ

, and any triangle with sides

a, b

and included angle

θ

has area

\frac{1}{2} ab sin θ

. Here

a = b = r

, giving

\frac{1}{2} r^{2} sin θ

Why is the centre not part of the circle itself?

The circle is the set of points at distance

r

from the centre; the centre is at distance

0

, so it fails the membership rule. It anchors the circle but sits inside it, not on it.

Why does the perpendicular-from-centre trick bisect a chord?

Take chord

A B

; drop a perpendicular from

O

meeting

A B

at the foot

M

(the red point below). This makes two right triangles

O M A

and

O M B

that share the side

O M

and have equal hypotenuses

O A = O B = r

; equal hypotenuse + equal leg + right angle forces them congruent, so

A M = M B

. This is the chord-bisection theorem.

Why does area grow with

r^{2}

while circumference grows with

r

Circumference measures a one-dimensional length (linear in

r

), whereas area fills a two-dimensional region — stretching in two directions at once, so it scales as

r \times r

. This squaring is why the ring-integration of

2 π r d r

lands on

π r^{2}

Why can the same chord define both a minor segment and a major segment?

The chord splits the disk into two pieces — the smaller region on the minor-arc side and the larger on the major-arc side. Both are segments; "minor/major" just names which side you mean.

Edge cases

The boundary and degenerate inputs — where sloppy formulas quietly break.

What is the segment area when

θ = 0

Zero. Plugging in gives

\frac{1}{2} r^{2} (0 - sin 0) = \frac{1}{2} r^{2} (0 - 0) = 0

— the chord has collapsed to a single point, enclosing no region, which matches.

What is the segment area when

θ = 2 π

(a full turn)?

The formula gives

\frac{1}{2} r^{2} (2 π - sin 2 π) = \frac{1}{2} r^{2} (2 π - 0) = π r^{2}

, the whole disk. That's correct: a "chord" spanning the full angle degenerates and the segment swallows the entire circle.

When

θ = π

(a diameter), what happens to the triangle term of the segment?

sin π = 0

, so the triangle area vanishes and the segment equals the sector:

\frac{1}{2} r^{2} π = \frac{1}{2} π r^{2}

, a clean semicircle. The chord is the diameter, so there's no triangle sticking out.

For a chord passing exactly through the centre, how far is it from the centre?

Distance

0

— it runs straight through

O

. This is the diameter, the limiting case where

r^{2} = A M^{2} + O M^{2}

with

O M = 0

, so the half-chord

A M = r

What is the length of the major arc for central angle

θ

The major arc covers the leftover angle

2 π - θ

, so its length is

r (2 π - θ) = 2 π r - r θ

. Check: minor

+

major

= r θ + (2 π r - r θ) = 2 π r

, the full circumference.

As the central angle

θ

approaches

0

, does the arc length approach the chord length?

Yes. For tiny

θ

the arc

r θ

and the chord

2 r sin (\frac{θ}{2})

both approach

r θ

, since

sin x \approx x

for small

x

. The curved edge and the straight cut become indistinguishable — the figure below shows the gap shrinking.

Can a "sector" ever have zero area while

r > 0

Yes, when

θ = 0

\frac{1}{2} r^{2} \cdot 0 = 0

. The two radii coincide, so no slice is enclosed even though the radius is positive.

Recall One-line survival kit

Same distance $r$ everywhere · diameter $d = 2 r$ (a chord through $O$ ) · $C = 2 π r$ · $A = π r^{2}$ · convert to radians before $s = r θ$ or $\frac{1}{2} r^{2} θ$ · minor arc $r θ$ , major arc $2 π r - r θ$ · segment $= \frac{1}{2} r^{2} (θ - sin θ)$ .