Circle Geometry. 108.03KB 2013-07-11 22:03:16

Maths Extension 1 – Circle Geometry

Circle Geometry
Properties of a Circle
Circle Theorems:
! Angles and chords
! Angles
! Chords
! Tangents
! Cyclic Quadrilaterals

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Maths Extension 1 – Circle Geometry

Properties of a Circle

Radius
Major Segment

Diameter

Chord
Minor Segment

Tangent

Concyclic points form a Cyclic Quadrilateral

Sector

Arc

Tangents Externally and Internally
Concentric circles

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Maths Extension 1 – Circle Geometry

Circle Theorems

! Equal arcs subtend equal angels at the
centre of the circle.

l

! If two arcs subtend equal angles at the
centre of the circle, then the arcs are
equal.

θ
O

l = rθ

θ

l

! Equal chords subtend equal angles at the
centre of the circle.

||

! Equal angles subtended at the centre of
the circle cut off equal chords.

θ
O

θ

||

A

B

||

θ

S OB = OC (radius of circle)
A ∠BOA = ∠COD (vert. opp. Angles)
S OA = OD (radius of circle)
∴ ∆BOA ≡ ∆COD (SAS)
AB = DC (corresponding sides in ≡ ∆ ' s )

O

θ

D

||
C

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Maths Extension 1 – Circle Geometry

! A perpendicular line from the centre of a

circle to a chord bisects the chord.
! A line from the centre of a circle that
bisects a chord is perpendicular to the
chord.
O

|

|

R ∠OMB = ∠OMA (straight line)
H OB = OA (radius of circle)
S OM = MO (common)
∴ ∆AOM ≡ ∆BOM (RHS)
AM = BM (corresponding sides in ≡ ∆ ' s )
O

A

|

M

|

B

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Maths Extension 1 – Circle Geometry

! Equal chords are equidistant from the
centre of the circle.
! Chords that are equidistant from the
centre are equal.

||

O

A

R ∠ANO = ∠BMO = 90° (A line from the
centre of a circle that bisects a chord is
perpendicular to the chord)
H AO = BO (Radius of Circle)
S NO = MO (given)
∴ ∆ANO ≡ ∆BMO (RHS)

N

O

||

M

B

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Maths Extension 1 – Circle Geometry

! The products of intercepts of intersecting
chords are equal

Internally
A

AX.XB = CX.XD
C

X

B

D

Prove ∆AXD ||| ∆CXD

A

C

X

A ∠AXD = ∠CXB (vertically opp)
A ∠XAD = ∠XCB (Angle standing on the
same arc)
A ∠XDA = ∠XBC (Angle sum of triangle)
Correspond sides
AX CX
=
XD XB
∴ AX . XB = CX . XD

D

B

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Maths Extension 1 – Circle Geometry

! The square of the length of the tangent
from an external point is equal to the
product of the intercepts of the secant
passing through this point.

Externally

(AX)2 = BX.CX
A

B

C

X

Externally

Prove ∆ACX ||| ∆BAX
A ∠AXC = ∠BXA (common)
A ∠XAC = ∠XBA (Angle in alternate
segment)
A ∠ACX = ∠BAX (Angle sum of triangle)
A

Correspond sides
AX BX
=
CX
AX
∴ ( AX ) 2 = BX .CX
B

C

X

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Maths Extension 1 – Circle Geometry

! The angle at the centre of a circle is twice
the angle at the circumference subtended
by the same arc.

θ

2θ

Let ∠CAO = α
Let ∠BAO = β
AO = CO = BO (radius of circle)
∴ ∠ACO = α (base angles of isosceles ∆ )
∴ ∠ABO = β (base angles of isosceles ∆ )
∠COD = 2α (exterior angle = two opposite
interior angles)
∠BOD = 2 β (exterior angle = two opposite
interior angles)

A

α β

O

β
α

B

C

∠CAB = α + β
∠COD = 2(a + β )

D

! Angle in a semicircle is a right angle.

180

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Maths Extension 1 – Circle Geometry

! Angles standing on the same arc are
equal.

θ

θ

Prove ∆AXD ||| ∆CXD

A

A ∠AXD = ∠CXB (vertically opp)
A ∠XAD = ∠XCB (Angle standing on the
same arc)
A ∠XDA = ∠XBC (Angle sum of triangle)
C

Corresponding angles of similar triangles are
equal

X
D

B

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Maths Extension 1 – Circle Geometry

! Tangents to a circle from an exterior
point are equal.

C

Prove ∆OAC ≡ ∆OBC
R ∠OAC = ∠OBC (90°)
H OC = OC (common)
S OA = )B (radii)
∴ ∆OAC ≡ ∆OBC

B
||

A

RHS

∴AC = BC (corresponding sides in
congruent triangles)

O

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Maths Extension 1 – Circle Geometry

! When two circles touch, the line through
their centres passes through their point of
contact.

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Maths Extension 1 – Circle Geometry

! The angle between a tangent and a chord
through the point of contact is equal to
the angle in the alternate segment.

θ

θ

Let ∠XAB = α
Let ∠BAQ = β
∴ α + β = 90 °

X
B
C

∠AXB = β (angle in semicircle is 90°,
complementary angle)
∠ACB = ∠AXB (angle on the same arc)
∴ ∠ACB = ∠BAQ
A

Q

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Maths Extension 1 – Circle Geometry

! The opposite angles in a cyclic
quadrilateral are supplementary.

A

α

! If the opposite angles of a quadrilateral
are supplementary, then the quadrilateral
is cyclic.
180 − β B

D

β
180 − α
C

! The exterior angle of s cyclic
quadrilateral is equal to the interior
opposite angle.

A

α

B

D

C

α

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