I'm Up and Down, and Round and Round: Circles

Contents hide

3Chords, Angles, and Distance from the Centre

5Concyclic Points and Cyclic Quadrilaterals

Chapter mind map: how it all connects

1 · Circles through pointsTwo points give infinite circles; three non-collinear points give exactly one: the circumcircle

2 · Chords and the centreEqual chords, equal angles at the centre, and equal distance from the centre, all imply each other

Circles

3 · Arcs and inscribed anglesAn arc’s angle at the centre is always double its angle anywhere else on the circle

4 · Concyclic pointsEqual angles on a chord, or opposite angles summing to 180°, both force four points onto one circle

What you will learn in this chapter

Why exactly one circle passes through any three non-collinear points
How equal chords, equal central angles, and equal distances from the centre are all connected
Why an arc’s angle at the centre is always twice its angle at the circle’s edge
What makes four points concyclic, and why opposite angles of a cyclic quadrilateral add to 180°

chorddiametercircumcirclecircumcentreconcycliccyclic quadrilateralarcinscribed angle

1Definitions and Symmetry

Figure 1: The parts of a circle: centre O, radius OQ, diameter PQ (a chord through the centre), and chord BC. — **Figure 1**: The parts of a circle: centre O, radius OQ, diameter PQ (a chord through the centre), and chord BC.

Learn by heartDefinition 1

A circle is the locus (set) of all points in a plane at a fixed distance, the radius, from a fixed point, the centre. A chord joins any two points on the circle; a diameter is a chord through the centre, the longest possible chord.

Symmetry

A circle has complete rotational symmetry about its centre (rotating by any angle leaves it unchanged) and reflection symmetry across every diameter.

2Circles Through Points

Through two points A, B, infinitely many circles pass, one for every point on the perpendicular bisector of AB chosen as centre (since every point on it is equidistant from A and B).

Figure 2: The circumcircle of triangle ABC: O is equidistant from A, B and C, so OA = OB = OC is the circumradius. — **Figure 2**: The circumcircle of triangle ABC: O is equidistant from A, B and C, so OA = OB = OC is the circumradius.

Theorem 1

A unique circle passes through any three non-collinear points

O lies on the perpendicular bisector of AB (since OA = OB is needed)any centre equidistant from A, B must be on that bisector

O also lies on the perpendicular bisector of AC (since OA = OC)same reasoning applied to A, C

Two non-parallel lines meet at exactly one point, so O is uniqueA, B, C non-collinear means these bisectors aren’t parallel

The circle centred at O with radius OA passes through A, B, Cby construction, OA = OB = OC

Where the circumcentre sits

For an acute triangle, the circumcentre O is inside the triangle. For an obtuse triangle, O is outside. For a right triangle, O sits exactly at the midpoint of the hypotenuse.

3Chords, Angles, and Distance from the Centre

Th 2

Equal chords → equal central angles

If AB = DE, then △CAB ≅ △CDE by SSS (all radii equal, given chords equal), so ∠ACB = ∠DCE.

Th 3

Equal central angles → equal chords

The converse: if ∠ACB = ∠DCE, then △ACB ≅ △DCE by SAS (two radii and the included angle), so AB = DE.

Th 4

Centre to midpoint ⊥ chord

△CAB is isosceles (CA = CB), so if M is the midpoint of AB, △CMA ≅ △CMB by SAS, giving ∠CMA = ∠CMB = 90°.

Th 5

Perpendicular from centre bisects chord

The converse of Theorem 4: drop a perpendicular from the centre to any chord, and it lands exactly at the midpoint.

Theorem 6

Chords of equal length are equidistant from the centre

Given AB = FG, with E, H the midpoints (feet of the perpendiculars from centre C)CE, CH are exactly the distances from C to the two chords

CA = CF and CB = CG (all radii), and AB = FG (given)setting up SSS for △CAB and △CFG

△CAB ≅ △CFG by SSS, so their altitudes from C are equal: CE = CHcongruent triangles have congruent corresponding altitudes

Theorem 7 (converse) and Theorem 8

Th 7: Chords equidistant from the centre are equal in length. Th 8: Of two unequal chords, the longer one is closer to the centre, proved using the Baudhāyana-Pythagoras theorem on the two right triangles formed by the radius, the half-chord, and the perpendicular distance.

Formula

Chord length

= 2\sqrt{r^2-d^2}

r

= radius,

d

= perpendicular distance from centre to chord. A diameter is the chord closest to the centre (d = 0) and the longest possible chord.

Given

Radius $r=13$ cm, chord is 5 cm from the centre

Find

Length of the chord

Solution

Chord $=2\sqrt{13^2-5^2}=2\sqrt{169-25}=2\sqrt{144}=2(12)=\;$ $24$ cm

4Angles Subtended by an Arc

Figure 3: ∠ACB (at the centre) is always double ∠ADB (at any point D on the major arc). — **Figure 3**: ∠ACB (at the centre) is always double ∠ADB (at any point D on the major arc).

Theorem 9

The angle an arc subtends at the centre is double the angle it subtends at any point on the circle outside the arc

Join D through C and extend to meet the circle at Ecreates two isosceles triangles to work with

△DCB isosceles (CB = CD), so ∠CBD = ∠CDB; exterior angle ∠BCE = 2∠BDCexterior angle of a triangle = sum of the two remote interior angles

Similarly △DCA gives exterior angle ∠ACE = 2∠ADCsame reasoning on the other side

∠BCA = ∠BCE + ∠ECA = 2(∠BDC + ∠CDA) = 2∠BDAadding the two results

Corollary: angle in a semicircle

If AB is a diameter, the arc from A to B (not through D) subtends a straight angle (180°) at the centre. So the angle it subtends at D is half that: ∠ADB = 90°, for any point D on the circle.

Angles in the same segment

Every point on the same arc sees a given chord at the same angle. This follows directly from Theorem 9, since all such points are “outside the arc” and share the same central angle.

Given

An arc subtends 70° at the centre

Find

The angle it subtends at a point on the circle

Solution

Angle at the circle $=\dfrac{70°}{2}=\;$ $35°$

5Concyclic Points and Cyclic Quadrilaterals

Learn by heartDefinition 2

Points that lie on the same circle are concyclic. A quadrilateral whose vertices are concyclic is a cyclic quadrilateral.

Theorem 10

If AB subtends equal angles at two points C, D on the same side of AB, then A, B, C, D are concyclic. Proved by contradiction: placing D off the circle through A, B, C forces an angle to be both equal to and strictly greater than itself.

Figure 4: Cyclic quadrilateral ABCD. ∠A + ∠C = 180° and ∠B + ∠D = 180°. — **Figure 4**: Cyclic quadrilateral ABCD. ∠A + ∠C = 180° and ∠B + ∠D = 180°.

Theorem 11

Opposite angles of a cyclic quadrilateral sum to 180°

∠BAD is half the (reflex) angle BOD, since A lies outside arc BCDTheorem 9 applied to arc BCD and point A

∠BCD is half the (non-reflex) angle BOD, since C lies outside arc BADTheorem 9 applied to arc BAD and point C

∠BAD + ∠BCD = half of (reflex BOD + BOD) = half of 360° = 180°the reflex and non-reflex angles at O make one full turn

Theorem 12 (converse) and a useful corollary

Th 12: If a quadrilateral’s opposite angles sum to 180°, it is cyclic (proved by contradiction, similar to Theorem 10). Corollary: the exterior angle at any vertex of a cyclic quadrilateral equals the interior opposite angle, since both are supplementary to the same interior angle.

Given

Cyclic quadrilateral PQRS with ∠P = (2x+10)°, ∠R = (3x−20)°

Find

$x$ , and the measures of ∠P and ∠R

Solution

Opposite angles: $(2x+10)+(3x-20)=180 \Rightarrow 5x-10=180 \Rightarrow x=38$

∠P $=2(38)+10=\;$ $86°$ , ∠R $=3(38)-20=\;$ $94°$

Every circle theorem in this chapter reduces to the same fact: any two radii are equal, which makes triangle after triangle isosceles.

the single idea to carry out of this chapter

Check yourself before the exam

Can I construct the circumcircle of a triangle and say where its centre lies?
Can I move between “equal chords”, “equal central angles” and “equidistant from centre”?
Can I state and apply “angle at centre = 2 × angle at circle”, including the 90°-in-a-semicircle corollary?
Can I prove that opposite angles of a cyclic quadrilateral sum to 180°, and use it in numericals?

Quick Revision: read this the night before the exam

Circle = locus equidistant from centre; diameter = longest chord
3 non-collinear points → unique circumcircle; centre = meeting point of perpendicular bisectors of the sides
Equal chords ⟺ equal central angles ⟺ equidistant from centre; longer chord is always closer to centre
Chord length $=2\sqrt{r^2-d^2}$
Angle at centre = 2 × angle at circle (same arc); angle in a semicircle = 90°
Cyclic quadrilateral: opposite angles sum to 180°; exterior angle = interior opposite angle

Practice Questions

2 marksA chord is 8 cm from the centre of a circle of radius 17 cm. Find the chord’s length.
1 markAn arc subtends 96° at the centre. What angle does it subtend at a point on the circle?
2 marksA circle of diameter 20 cm has a chord of length 16 cm. Find its distance from the centre.
3 marksProve that the perpendicular bisector of any chord of a circle always passes through the centre.
1 markAB is a diameter of a circle and C is any point on the circle. What is ∠ACB? Why?
2 marksIn cyclic quadrilateral WXYZ, ∠W = 85°. Find ∠Y. If ∠X = 100°, find ∠Z.
3 marksQuadrilateral EFGH is cyclic with ∠E = (3x+5)° and ∠G = (2x+15)°. Find x, ∠E and ∠G.
2 marksA chord of length 24 cm is 5 cm from the centre. Find the circle’s radius.
4 marksTwo parallel chords of length 10 cm and 24 cm lie on opposite sides of the centre of a circle of radius 13 cm. Find the distance between them.
3 marksExplain, with reasons, why no chord of a circle can ever be longer than the diameter.

Multiple Choice

1 markThe circumcentre of an obtuse-angled triangle lies:
(a) inside the triangle(b) outside the triangle(c) at the midpoint of a side(d) at a vertex
1 markIn a cyclic quadrilateral, if one angle is 65°, its opposite angle is:
(a) 65°(b) 115°(c) 25°(d) 180°
1 markThe angle in a semicircle is always:
(a) 45°(b) 60°(c) 90°(d) 180°

Assertion (A): Two chords of equal length in the same circle are equidistant from the centre.
Reason (R): Equal chords subtend equal angles at the centre, and this forces their perpendicular distances from the centre to be equal too.

Case Study

A circular park has four gates A, B, C, D placed on its boundary, forming a quadrilateral ABCD. A surveyor measures ∠A = 78° and ∠B = 95°.

1 markWhy must ABCD be a cyclic quadrilateral?
1 markFind ∠C.
2 marksFind ∠D.

Answer Key

2\sqrt{17^2-8^2}=2\sqrt{289-64}=2\sqrt{225}=30

96/2=48°

Q3radius

=10

;

\sqrt{10^2-8^2}=\sqrt{36}=6

Q4Follows from Theorem 4/5: the centre is equidistant from the chord’s endpoints, so it must lie on their perpendicular bisector

90°

, angle in a semicircle (Theorem 9 corollary)

Q6∠Y

=180-85=95°

; ∠Z

=180-100=80°

(3x+5)+(2x+15)=180 \Rightarrow x=32

; ∠E

=101°

, ∠G

=79°

r=\sqrt{12^2+5^2}=\sqrt{169}=13

Q9Half-chords 5, 12; distances from centre

\sqrt{13^2-5^2}=12

and

\sqrt{13^2-12^2}=5

; opposite sides, so total distance

=12+5=17

Q10Any chord not through the centre forms a right triangle with the radius as hypotenuse, so it is shorter than 2r; the diameter (through the centre) attains the maximum, 2r

MCQ 1–3(b) · (b) · (c)

A/RBoth A and R are true, and R is the correct explanation of A

Case StudyAll four gates lie on the park’s circular boundary, so A, B, C, D are concyclic, which is exactly what makes ABCD a cyclic quadrilateral · ∠C

=180°-78°=102°

· ∠D

=180°-95°=85°