Download radii: AP , PR,PB diameter: AB chords: AB , CD, AF secant: AG or AG

L2 Ch 6 Note Sheet Key
Review: Circles Vocabulary
Name ___________________________
If you are having problems recalling the vocabulary, look back at your notes for Lesson 1.7 and/or page 69 – 71 of
your book. Also, pay close attention to the geometry notation you need to use to name the parts!!
Circles
G
Circle is a set of points in a plane a given distance
(radius) from a given point (center).
F
R
Congruent circles are two or more circles with the
same radius measure.
Concentric circles are two or more circles with the
same center point.
A
C
P
B
D
Types of Lines [Segments]
Radius [of a circle] is
• A segment that goes from the center to any point
on the circle.
• the distance from the center to any point on the
circle.
Diameter [of a circle] is
• A chord that goes through the center of a circle.
• the length of the diameter. d = 2r or ½ d = r.
Chord is a segment connecting any two points on the
circle.
Secant A line that intersects a circle in two points.
Tangent [to a circle] is a line that intersects a circle in
only one point. The point of intersection is called the
point of tangency.
Arcs & Angles
Arc [of a circle] is formed by two points on a circle and a
continuous part of the circle between them. The two
points are called endpoints.
Semicircle is an arc whose endpoints are the endpoints of
the diameter.
Minor arc is an arc that is smaller than a semicircle.
Major arc is an arc that is larger than a semicircle.
Intercepted Arc An arc that lies in the interior of an
angle with endpoints on the sides of the angle.
Central angle An angle whose vertex lies on the center
of a circle and whose sides are radii of the circle.
Measure of a central angle determines the measure of
an intercepted minor arc.
S. Stirling
E
radii: AP , PR , PB
diameter: AB
chords: AB , CD , AF
secant: AG or AG
tangent: EB
semicircles: ACB , ARB
minor arcs: AC , AR , RD , BC ,…
major arcs: ARC , ARD , FRD ,
BAC ,…
central angles with their intercepted
arcs:
m∠APR = mAR
m∠RPB = mRB
Page 1 of 10
L2 Ch 6 Note Sheet Key
Lesson 6.1 Tangent Properties
Investigation 1: See Worksheet page 1.
Tangent Conjecture
A tangent to a circle is
perpendicular to
the radius drawn to the
point of tangency.
Name ___________________________
Converse of the Tangent
Conjecture
A line that is perpendicular
to a radius at its endpoint
on the circle is tangent to
the circle.
O
T
N
Investigation 2: See Worksheet page 1.
Tangent Segments Conjecture
Tangent segments to a circle from a point
outside the circle are congruent.
A
E
G
Vocabulary (review):
Intercepted Arc An arc that lies in the interior of an angle with
endpoints on the sides of the angle.
G
Central angle
∠GNA determines the measure of minor arc GA .
∠GNA is said to intercept GA because the arc GA is within
the angle.
N
T
A
M
Are there any relationships between the angles formed when you
have tangent segments from a point outside a circle?
If mGA = 120° …
Minor arc = central angle, m∠GNA = mGA = 120° .
TA and TG are tangent to circle N, so m∠A and m∠G both = 90°.
The sum of the angles of quadrilateral TANG must be 360°. So m∠T + 90 + 140 + 90 = 360° , which
means m∠T = 60° .
What is the relationship between ∠T and the intercepted arc, GA ? They add to 180°
S. Stirling
Page 2 of 10
L2 Ch 6 Note Sheet Key
Name ___________________________
Intersecting Tangents Conjecture
The measure of an angle formed by two intersecting tangents
to a circle is 180 minus the measure of the intercepted arc,
x = 180 − a .
Example 1
180 = x + 110 , x = 70
x
a
Example 2
Rays r as s are tangents. w = ?
180 – 54 = w
OR
so central angle = w
360 – 90 – 90 – 54 = w
126 = w
y = 5 cm Tangents to a circle from a point are congruent.
mNP = 110 Measure of a central angle equals its intercepted
arc.
mPQN = 360 − 110 = 250 Total degrees in a circle = 360°
Example 3
AD is tangent to both circle B and circle C.
w = 100
mAXT = 260
m∠A = 90 and m∠D = 90 , w = 360 − 90 − 90 − 80 = 100
w = mAT = 100 , mAXT = 360 − 100 = 260
S. Stirling
Page 3 of 10
b
L2 Ch 6 Note Sheet Key
Name ___________________________
Vocabulary:
“Tangent” means to intersect in one point.
Tangent Circles Circles that are tangent to the same line at the same point.
Internally Tangent
Circles
Two tangent circles
having centers on the
same side of their
common tangent.
Externally Tangent
Circles
Two tangent circles
having centers on
opposite sides of their
common tangent.
Polygons and Circles
The triangle is inscribed in the circle, or
the circle is circumscribed about the triangle.
The triangle is circumscribed about the circle, or
the circle is inscribed in the triangle.
B
All of the vertices
of the polygon are
on the circle.
The sides are all
chords of the circle.
P
All of the sides of the polygon
are tangent to the circle.
C
B
O
C
A
A
Lesson 6.2 Chord Properties
Read top of page 317. Then use the
diagrams to define the following.
Examples:
Non-Examples:
R
D
P
Central angle
An angle whose vertex lies on the
center of a circle and whose sides
are radii of the circle.
Measure of a central angle =
measure of its intercepted arc.
O
Q
A
S
T
B
∠AOB , ∠DOA and
∠DOB are central
∠PQR , ∠PQS , ∠RST ,
∠QST and ∠QSR are NOT
angles of circle O.
central angles of circle O.
Examples:
Non-Examples:
Q
A
Inscribed angle
An angle whose vertex lies on a
circle and whose sides are chords of
the circle.
C
B
E
R
T
D
∠ABC , ∠BCD and
∠CDE are inscribed
angles.
S. Stirling
V
P
W
X
S
U
∠PQR , ∠STU , and ∠VWX
are NOT inscribed angles.
Page 4 of 10
L2 Ch 6 Note Sheet Key
Name ___________________________
Investigation 3: See Worksheet page 2.
Example 1
Chord Conjectures
If two chords in a circle are congruent, then
• they determine two central angles that
are congruent.
• their intercepted arcs are congruent.
• are equidistant from the center of the
circle.
B
D
If AB ≅ CD , then
O
∠BOA ≅ ∠COD .
A
If AB ≅ CD , then
C
AB ≅ CD
Example 2
B
D
O
If AB ≅ CD , then
M
OM = ON .
N
A
Investigation 4: See Worksheet page 2.
Perpendicular to a Chord Conjecture
The perpendicular from the center of a circle
to a chord bisects the chord and its
intercepted arc.
Perpendicular Bisector of a Chord
Conjecture
The perpendicular bisector of a chord passes
through the center of the circle.
Note: This property is used to find the center of any
circle. Just draw two nonparallel chords, of any
length, then construct the perpendicular bisectors of
both of them. By the Perpendicular Bisector of a
Chord Theorem, the intersection must be the center
of the circle.
S. Stirling
Example 3
C
F
If OM ⊥ EF , then
O
Q
OM bisects EF
or EM ≅ MF
and EQ ≅ QF
M
E
Example 4
j
If line j is the
perpendicular bisector
of BD , then
line j passes through
point P.
A
D
B
P
k
C
If line k is the perpendicular bisector of AC
then line k passes through point P.
Page 5 of 10
L2 Ch 6 Note Sheet Key
Lesson 6.3 Arcs and Angles
Read top of page 324.
Investigation 5: See Worksheet page 5.
Inscribed Angle Conjecture
The measure of an angle inscribed in a circle
equals half the measure of its intercepted arc.
Name ___________________________
Example 1
C
If m∠COR = 92° ,
then m∠CAR = ? .
92
O
A
R
mCR = 92° .
The central angle equals its intercepted arc.
If mCR = 92° , then m∠CAR = 46° .
The inscribed angle equals half its
intercepted arc.
Example 2
mAB = 170° , find m∠APB and m∠AQB .
A
P
m∠APB = 1 (170 ) ° = 85°
2
m∠AQB = 1 (170 ) ° = 85°
2
170
Q
or The inscribed angles intercept the same arcs, so they’re congruent.
B
Example 3
mAB = mCD = 42° , find m∠APB and m∠CQD
m∠APB = m∠CQD = 1 i42 = 21° .
2
The inscribed angle equals half its intercepted arc.
A
P
42
B
O
C
42
Q
D
R
Example 4
∠ACD , ∠ABD and ∠AED intercept semicircle ARD .
A
Find the measure of each angle.
All are 1 i180 = 90°
2
m∠ACD = m∠ABD = m∠AED = 90°
.
S. Stirling
O
C
B
Page 6 of 10
D
E
L2 Ch 6 Note Sheet Key
Name ___________________________
Investigation 6: See Worksheet page 5.
Cyclic Quadrilateral A quadrilateral that
can be inscribed in a circle.
Each of its angles are inscribed angles.
Each of its sides is a chord of the circle.
If ABDC is a cyclic quadrilateral
m∠A + m∠D = 180° and m∠B + m∠C = 180°
A
Cyclic Quadrilateral Conjecture
The opposite angles of a cyclic
quadrilateral are supplementary.
C
m ∠BAC = 106.61°
m ∠CDB = 73.39°
m ∠DBA = 85.99°
B
m ∠ACD = 94.01°
D
G
Cyclic Parallelogram Conjecture
If a parallelogram is inscribed within a circle, then the
parallelogram is a rectangle.
O
Y
D
L
Secant
A line that intersects a circle in two
points.
Secant Segment
A segment that intersects a circle in
two points and it contains at least
one point on the exterior of the
circle.
Investigation 7: See Worksheet page 6.
Parallel Lines [Secants] Intercepted
Arcs Conjecture
Parallel lines [secants] intercept
congruent arcs on a circle.
S. Stirling
Secants: EC , TN
L
Secant Segments: LN , SA ,
T
SC , EA .
Chords: EC , TN because they
begin and end on the circle.
N
S
A
C
E
If AB DC ,
then mAD = mBC = 64° .
D
64
A
Note: mAB ≠ mDC .
C
B
Page 7 of 10
64
L2 Ch 6 Note Sheet Key
Name ___________________________
Are there any short cuts for finding the measures of angles and their intercepted arcs?
Quick Formulas for Angles and Arcs:
Diagram:
Where is the What is the
vertex of the figure formed
angle?
by?
Center of the 2 radii
A
circle
“Central
O 1
angle”
Measure of the angle formed?
angle = intercepted arc
m∠1 = m AB
B
On the circle
2 chords
“Inscribed
angle”
angle = ½ (intercepted arc)
C
E
m∠ 2 = 1 i mCE
2
2
D
1 chord &
1 tangent
m∠ 2 = 1 i m FG
2
F
2
H
Inside the
circle
G
2 chords
angle = ½ (sum of intercepted arcs)
L
m ∠3 = 1
M
3
V
2
( m JK + mLM )
J
K
Outside the
circle
2 secants
angle
= ½ (difference of intercepted arcs)
N
O
4
P
1 secant &
1 tangent
W
U
T
R
Q
4
( m NP − mOQ )
m∠ 4 = 1 ( mWS − m SU )
2
m∠ 4 = 1
2
S
2 tangents
X
4
Y
Summary:
Vertex center
m∠ 4 = 1
Z
2
( mYAZ − mYZ )
A
○ → angle = arc
○ → angle = ½ arc
Vertex inside ○ → angle = ½ sum arcs
Vertex outside ○ → angle = ½ difference arcs
Vertex on
S. Stirling
Page 8 of 10
L2 Ch 6 Note Sheet Key
Name ___________________________
Lesson 6.5 The Circumference/Diameter Ratio Read top of page 335.
Pi is a number, just like 3 is a number. It represents the number you get when you take the circumference of a
circle and divide it by the diameter. There is no exact value for pi, so you use the symbol π. You will leave π in the
answers to your problems unless they ask for an approximate answer. If they do, use 3.14 or the π key on your
calculator.
Circumference
The perimeter of a circle, which is the distance around the circle. Also, the curved path of
the circle itself.
Circumference Conjecture
If C is the circumference and d is the diameter of a circle, then there is a number π such
that C = dπ
Since d = 2r, where r is the radius, then C = 2rπ or 2πr
Example A: If a circle has a diameter of 3 meters,
what is it’s circumference?
Example B: If a circle has a circumference
of 12π meters, what is it’s radius?
C =πd
C = π •3
C = 3π
C ≈ 9.4 m
C = 2π r
12π = 2π r
12π 2π r
=
2π
2π
r = 6m
write the formula
substitute
simplify (pi written last, like a variable)
Only an approximate answer!
write the formula
substitute
divide to get r.
simplify
Example C:
If a circle has a circumference of 20 meters, what
is it’s diameter?
C =πd
20 = π d
20
=d
π
write the formula
substitute
accurate answer
d ≈ 6.366 m approximate
S. Stirling
Page 9 of 10
L2 Ch 6 Note Sheet Key
Lesson 6.7 Arc Length
Investigation 10: See Worksheet page 11.
Name ___________________________
Measure of an Arc: The measure of an arc equals the measure of its central angle,
measured in degrees.
Arc length: The portion of (or fraction of) the circumference of the circle described by an
arc, measured in units of length.
Arc Length Conjecture
The length of an arc equals the measure of the arc divided by 360° times the
circumference.
It is a fraction of the circle! So…….
Use the formula every time!! Arc Length =
arc degrees
• Circumference
360
Example (a):
Given central ∠ATB = 90°
with radius 12.
Example (b):
Given central ∠COD = 180°
with diameter 15 cm.
mAB = 90°
mCED = 180°
90 1
=
fraction of circle =
360 4
90
length of AB =
( 2π i12)
360
1
= ( 24π ) = 6π
4
180 1
=
fraction of circle =
360 2
180
length of CED =
(15π )
360
15
= π cm
2
EXAMPLE B:
If the radius of a circle is 24 cm and
m∠BTA = 60° , what is the length of AB ?
mAB = 120°
T
S. Stirling
fraction of circle =
120 1
=
360 3
1
( 2π i9)
3
= 6π ft
length of EF =
The length of ROT is 116π, what is the radius of
the circle?
length of AB =
60
Given mEF = 120° with
radius = 9 ft.
EXAMPLE C:
B
120
( 2π i24)
360
= 16π ≈ 50.3 cm
Example (c):
A
mROT = 360 − 120° = 240
240
116π =
( 2π ir )
360
4π
116π =
r
3
116π 3
i = r = 87 .
then
1 4π
O
R
Page 10 of 10
120
T