Parts of a Circle – GCSE Maths

1 The Vocabulary of the Circle

Circle questions are full of specific words. Getting them right is half the battle, because each one brings a property with it.

Word	Meaning
Centre	The point in the middle, usually labelled $O$.
Radius	A line from the centre to the edge. Plural: radii.
Diameter	A line right across the circle, through the centre. $d = 2r$.
Circumference	The distance all the way round the outside — the circle's perimeter.
Chord	A straight line joining two points on the circle (not through the centre).
Arc	Part of the circumference. The minor arc is the shorter one, the major arc the longer.
Sector	A "pizza slice" — the region between two radii and an arc.
Segment	The region between a chord and an arc.
Tangent	A straight line that touches the circle at exactly one point.
Semicircle	Half a circle, cut off by a diameter.

Sector or segment? A sector is bounded by two straight radii (like a slice of pizza). A segment is bounded by one straight chord (like the piece you cut off the top of an orange). Mixing these up is the single most common circle vocabulary error.

2 The Basic Properties

The property you use most

All radii of a circle are equal.

That single fact is why circles produce so many isosceles triangles. Any triangle formed by two radii and a chord automatically has two equal sides, so its base angles are equal. Spotting this unlocks a huge number of questions.

Property	Why it is useful
All radii are equal	Creates isosceles triangles everywhere
The diameter is twice the radius	$d = 2r$, $r = \tfrac{d}{2}$
The diameter is the longest chord	Useful in "greatest distance" problems
A tangent meets a radius at $90^\circ$	Creates right-angled triangles for Pythagoras
A radius perpendicular to a chord bisects it	Splits the chord into two equal halves
Two tangents from the same external point are equal	Creates a kite and an isosceles triangle

Worked Example 1 — Using the isosceles property

$A$ and $B$ are points on a circle with centre $O$. Angle $AOB = 44^\circ$. Find angle $OAB$.

①$OA = OB$ (radii of the same circle), so triangle $OAB$ is isosceles.

②The base angles $OAB$ and $OBA$ are therefore equal.

③$180 - 44 = 136^\circ$ shared between them.

④$\angle OAB = 136 \div 2 = 68^\circ$

3 Tangents

Tangent property 1

A tangent is perpendicular to the radius drawn to the point of contact.

Tangent property 2

The two tangents drawn from an external point $P$ are equal in length: $PA = PB$.

Because $PA = PB$ and $OA = OB$, the shape $OAPB$ is a kite. The line $OP$ bisects both $\angle APB$ and $\angle AOB$.

Worked Example 2 — Tangent and Pythagoras

A circle has centre $O$ and radius $5$ cm. $P$ is a point $13$ cm from $O$. A tangent from $P$ touches the circle at $T$. Find $PT$.

①$OT$ is a radius and $PT$ is a tangent, so $\angle OTP = 90^\circ$.

②Triangle $OTP$ is right-angled with hypotenuse $OP = 13$.

③$PT^2 = 13^2 - 5^2 = 169 - 25 = 144$

④$PT = \sqrt{144} = 12$ cm

Worked Example 3 — Angles in a tangent kite

Two tangents from $P$ touch a circle, centre $O$, at $A$ and $B$. Angle $APB = 54^\circ$. Find angle $AOB$.

①$\angle OAP = \angle OBP = 90^\circ$ (tangent perpendicular to radius)

②$OAPB$ is a quadrilateral, so its angles add to $360^\circ$.

③$\angle AOB = 360 - 90 - 90 - 54 = 126^\circ$

Notice the shortcut: $\angle AOB$ and $\angle APB$ always add to $180^\circ$ in this configuration.

4 Chords

The chord property

The perpendicular from the centre to a chord bisects the chord.

Why? Join the centre $O$ to both ends of the chord $AB$, and let $M$ be the foot of the perpendicular. Then $OA = OB$ (radii), $OM = OM$ (common) and both triangles have a right angle at $M$. So $\triangle OAM \cong \triangle OBM$ by RHS, giving $AM = MB$.

Worked Example 4 — Finding a chord length

A circle has radius $13$ cm. A chord is $5$ cm from the centre. Find the length of the chord.

①Draw the perpendicular from $O$ to the chord, meeting it at $M$. Then $OM = 5$ cm.

②Join $O$ to one end $A$: $OA = 13$ cm (a radius) and $\angle OMA = 90^\circ$.

③$AM^2 = 13^2 - 5^2 = 169 - 25 = 144$, so $AM = 12$ cm.

④$M$ is the midpoint, so the whole chord $AB = 2 \times 12 = 24$ cm.

Don't stop at $12$! Doubling at the end is the step most often forgotten.

Worked Example 5 — Finding the distance from the centre

A chord of length $16$ cm is drawn in a circle of radius $10$ cm. How far is the chord from the centre?

①Half the chord $= 16 \div 2 = 8$ cm.

②Right-angled triangle: hypotenuse $10$ (radius), one leg $8$.

③$OM^2 = 10^2 - 8^2 = 100 - 64 = 36$

④$OM = 6$ cm

Equal chords are equidistant from the centre, and conversely, chords the same distance from the centre are equal in length. The closer a chord is to the centre, the longer it is — which is why the diameter, at distance $0$, is the longest chord of all.

5 Quick Reference

Radius / diameter

$d = 2r$. All radii of a circle are equal.

Chord

Joins two points on the circle. The diameter is the longest chord.

Arc

Part of the circumference; minor is shorter, major is longer.

Sector

Bounded by two radii and an arc — a pizza slice.

Segment

Bounded by a chord and an arc.

Tangent

Touches at one point; perpendicular to the radius there.

Two tangents

From the same external point they are equal, forming a kite.

Chord bisector

The perpendicular from the centre cuts a chord in half.

Look for

Two radii and a chord always make an isosceles triangle.

6 Practice Questions

Question 1

Name the part of the circle described in each case: (a) a straight line joining two points on the circle; (b) the region between two radii and an arc; (c) a line touching the circle at exactly one point.

▶ Show solution

(a) A chord (a diameter if it passes through the centre).

(b) A sector.

(c) A tangent.

Question 2

A circle has diameter $17$ cm. Write down its radius.

▶ Show solution

$r = \dfrac{d}{2} = \dfrac{17}{2} = 8.5$ cm

Question 3

$A$ and $B$ lie on a circle with centre $O$, and angle $AOB = 96^\circ$. Find angle $OBA$, giving a reason.

▶ Show solution

$OA = OB$ (radii of the same circle), so triangle $OAB$ is isosceles.

Base angles are equal: $\angle OAB = \angle OBA$.

$(180 - 96) \div 2 = 84 \div 2 = 42^\circ$

Question 4

A tangent from an external point $P$ touches a circle of radius $8$ cm at $T$. Given $PT = 15$ cm, find the distance $OP$.

▶ Show solution

$\angle OTP = 90^\circ$ (tangent perpendicular to radius).

$OP^2 = 8^2 + 15^2 = 64 + 225 = 289$

$OP = \sqrt{289} = 17$ cm

Question 5

A chord of a circle of radius $25$ cm is $7$ cm from the centre. Find the length of the chord.

▶ Show solution

Half-chord$^2 = 25^2 - 7^2 = 625 - 49 = 576$

Half-chord $= \sqrt{576} = 24$ cm

Full chord $= 2 \times 24 = 48$ cm

Question 6

Explain the difference between a sector and a segment.

▶ Show solution

A sector is bounded by two radii and an arc — the shape of a slice of pizza, with its point at the centre.

A segment is bounded by one chord and an arc — the shape you cut off when you slice straight across a circle without going through the centre.

Question 7

Two tangents from a point $P$ touch a circle with centre $O$ at $A$ and $B$. Angle $AOB = 140^\circ$. Find angle $APB$.

▶ Show solution

$\angle OAP = \angle OBP = 90^\circ$ (tangent perpendicular to radius).

Angles of quadrilateral $OAPB$ add to $360^\circ$:

$\angle APB = 360 - 90 - 90 - 140 = 40^\circ$

Question 8

In a circle of radius $10$ cm, chord $PQ$ has length $12$ cm and chord $RS$ has length $16$ cm. Which chord is closer to the centre? Justify with calculation.

▶ Show solution

$PQ$: half-chord $= 6$; distance$^2 = 100 - 36 = 64$; distance $= 8$ cm.

$RS$: half-chord $= 8$; distance$^2 = 100 - 64 = 36$; distance $= 6$ cm.

$RS$ is closer ($6$ cm vs $8$ cm) — which fits the rule that the longer chord is always nearer the centre.

Question 9

Two tangents from $P$ touch a circle of radius $9$ cm at $A$ and $B$. The distance $OP = 41$ cm. Find (a) the length $PA$, (b) the perimeter of the kite $OAPB$.

▶ Show solution

(a) $\angle OAP = 90^\circ$, so $PA^2 = 41^2 - 9^2 = 1681 - 81 = 1600$.

$PA = 40$ cm.

(b) $PB = PA = 40$ cm (equal tangents) and $OA = OB = 9$ cm (radii).

Perimeter $= 9 + 40 + 40 + 9 = 98$ cm.

Question 10

A circular tabletop of radius $60$ cm has a straight edge cut off along a chord that is $36$ cm from the centre.

(a) Find the length of the straight edge. (b) Find the greatest width of the piece cut off (measured perpendicular to the chord). (c) What is the greatest distance between any two points on the remaining tabletop?

▶ Show solution

(a) Half-chord$^2 = 60^2 - 36^2 = 3600 - 1296 = 2304$.

Half-chord $= \sqrt{2304} = 48$ cm, so the straight edge is $2 \times 48 = \mathbf{96}$ cm.

(b) The piece cut off is a segment. Its greatest width is the distance from the chord out to the circle, along the line through the centre:

$60 - 36 = \mathbf{24}$ cm.

(c) The full diameter is $120$ cm. Is it still present? The chord is $36$ cm from the centre, so the centre is still on the tabletop, and a diameter drawn parallel to the chord is untouched by the cut.

So the greatest distance is still the full diameter, $\mathbf{120}$ cm.

Parts of a Circle (G9) · GCSE Maths Revision · Created with MathJax