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New Jersey Center for Teaching and Learning
Progressive Mathematics Initiative®
This material is made freely available at www.njctl.org
and is intended for the non-commercial use of
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community, and/or provide access to course
materials to parents, students and others.
Geometry
Circles
2014-06-03
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Click to go to website:
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Table of Contents
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Click on a topic to go
to that section
Parts of a Circle
Central Angles & Arcs
Arc Length & Radians
Chords, Inscribed Angles & Triangles
Circles
and
Their Parts
Tangents & Secants
Return to the
table of
contents
Segments & Circles
Questions from Released
PARCC Examination
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Circles are a type of Figure
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Circles
A figure lies in a plane and is contained by a boundary.
Euclid defined a circle and its center in this way:
Euclid defined figures in this way:
Definition 15: A circle is a plane figure contained by one line
such that all the straight lines falling upon it from one point
among those lying within the figure equal one another.
Definition 13: A boundary is that which is an extremity of anything.
Definition 14: A figure is that which is contained by any boundary or
boundaries.
A boundary divides a plane into those parts that are within the
boundary and those parts that are outside it. That which is within the
boundary is the "figure."
Definition 16: And the point is called the center of the circle.
This states that all the radii (plural of radius) drawn from the
center of a circle are of equal length, which is a very
important aspect of circles and their radii.
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Circles and Their Parts
Another way of saying this is that a circle is made up of all
the points that are an equal distance from the center of the
circle.
Circles and Their Parts
The symbol for a circle is
and is named by a capital letter
placed by the center of the circle.
The below circle is named: Circle A or
radius
center
AB is a radius of
A
A radius (plural, radii) is a line
segment drawn from the center of the
circle to any point on the circle.
A
circumference
A
B
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Radii
A radius (plural, radii) is a line segment drawn from the center
of the circle to any point on its circumference.
It follows from the definition of a circle that all the radii of a
circle are congruent since they must all have equal length.
Diameters
Definition 17: A diameter of the circle is any straight line drawn
through the center and terminated in both directions by the
circumference of the circle, and such a straight line also bisects
the circle.
An unlimited number of radii can be drawn in a circle.
C
That all radii of a circle are
congruent will be important to
solving problems.
A
In this drawing, we know that
line segments AC, AD and AB
are all congruent.
B
D
Since the diameter passes through the center of the circle and
extends to the circumference on either side, it is twice the length
of a radius of that circle.
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Diameters
There are an unlimited number of diameters which can be drawn
within a circle.
They are all the same length, so they are all congruent.
Chords
A chord is a line segment whose endpoints lie on the
circumference of the circle.
So, a diameter is a special case of a chord.
Why is a radius not a chord?
C
That all diameters of a circle
are congruent will be important
to solving problems.
F
E
A
G
D
B
In this drawing, we know that
line segments BE, CG and DF
are all congruent.
F B
G
D
All the line segments in this drawing
are chords.
A
E
C
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Semicircles
Chords
There are an unlimited number of chords which can be
drawn in a circle.
Definition 18: A semicircle is the figure contained by the
diameter and the circumference cut off by it. And the center of
the semicircle is the same as that of the circle.
Chords are not necessarily the same length, so are not
necessarily congruent.
F B
diameter
D
Chords can be of any length up to a
maximum.
A
G
semicircles
What is the longest chord that can be
drawn in a circle?
E
C
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Diameters and Radii
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1
A diameter of a circle is the longest chord of the
circle.
The measure of the diameter, d, is twice the measure of
the radius, r.
True
In this case, CD = 2 AB
False
In general, d = 2r
or
r = 1/2 d
Example: In the diagram to the
left, AB = 5. Determine AC & DC.
D
A
B
C
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2
A radius of a circle is a chord of a circle.
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3
The length of the diameter of a circle is equal to
twice the length of its radius.
True
False
True
False
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4
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If the radius of a circle measures 3.8 meters, what
is the measure of the diameter?
5
How many diameters can be drawn in a circle?
A
1
B
2
C
4
D
infinitely many
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Arcs
A
Central Angles
& Arcs
Arc AB or AB
B
Arc AB is the path between points A
and B on the circumference of the
circle.
Return to the
table of
contents
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Arcs
A
Arc AB or AB
B
C
Arcs
A
Of course, you may have wondered
why we went around the circle the
way we did (in blue).
This shorter path (blue) is called the
minor arc. We use 2 letters to
denote minor arcs
The longer path (green) is called the
major arc. If we want to refer to a
major arc, we will add another point
along that path and include it in the
name.
For instance, we would name the
green arc, Arc ACB to distinguish it
from Arc AB.
Arc AB or AB
B
Now let's discuss some ways we
can use arcs, and measure them.
C
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Central Angles
Central Angles
A
A
B
O
A central angle of a circle is any
angle which has vertices consisting
of the center of the circle and two
points on the circumference.
O
B
700
So, the measure of the angle AOB is
the same as the measure of Arc AB,
also denoted by mAB.
How many different central angles
can you find in this diagram?
D
C
C
Name them.
The measure of a central angle is
equal to the measure of the arc it
intercepts.
In this case, the mAB is 700, since
that is the m∠AOB.
That also means the mACB is 2900,
since a full trip around the circle is
3600.
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Adjacent Arcs
A
C
Adjacent Arcs
A
Adjacent arcs: two arcs of
the same circle are
adjacent if they have a
common endpoint.
T
O
From the Angle Addition Postulate
we know that
m∠COT = m∠COA + m∠AOT
In this case, Arc CA and
Arc AT are adjacent since
they share the endpoint A.
C
T
O
It follows then that the measures
of adjacent arcs can be added to
find the measure of the arc formed
by the adjacent arcs.
In this case that:
mCAT = mCA + mAT
which is the Arc Addition Postulate
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6 Find the measure of Arc RU.
7 Find the measure of Arc RT.
T
T
300
1000
600
R
900
800
V
300
U
1000
600
R
900
800
V
U
Answer
S
Answer
S
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8 Find the measure of Arc RVT.
9 Find the measure of Arc UST.
T
T
S
900
1000
U
80
0
600
R
300
Answer
300
900
1000
600
V
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10 Which type of arc is Arc TQR?
11 Which type of arc is Arc QRT?
A Minor Arc
A Minor Arc
B Major Arc
B Major Arc
C Semicircle
C Semicircle
D None of these
D None of these
T
Q
Q
120
600
120
600
0
0
800
S
800
R
Note that you need to use the indicated
degree measures as the drawing is not
to scale.
S
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R
Note that you need to use the indicated
degree measures as the drawing is not
to scale.
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12 Which type of arc is Arc QS?
13 Which type of arc is Arc TS?
A Minor Arc
B Major Arc
U
800
R
V
T
Answer
S
A Minor Arc
T
Q
120
600
0
C Semicircle
B Major Arc
T
Q
120
600
0
C Semicircle
800
D None of these
800
D None of these
S
Note that you need to use the indicated
degree measures as the drawing is not
to scale.
R
S
Note that you need to use the indicated
degree measures as the drawing is not
to scale.
R
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Congruent Circles and Arcs
14 Which type of arc is Arc RST?
A Minor Arc
· All circles are similar since they all have the identical shape
B Major Arc
C Semicircle
· Circles which have the same radius are congruent, since
they will overlap at every point if their centers are lined up.
D None of these
T
Q
120
600
0
800
Note that you need to use the indicated
degree measures as the drawing is not
to scale.
R
S
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Congruent Circles and Arcs
· Arcs are similar if they have the same measure
· Arcs are congruent if they have the same measure and
are either part of the same circle or of congruent circles.
D
C
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Congruent Circles and Arcs
Since the measure of an arc is equal to that of the
central angle which intercepts it, arcs within the same
circle which are intercepted by central angles of the
same measure, are congruent.
D
E
550
550
O
F
Arc CE and Arc EF are congruent because they are in the
same circle and have the same measure.
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Congruent Circles and Arcs
· Arcs are similar if they have the same measure
· Arcs are congruent if they have the same measure and
are either part of the same circle or of congruent circles.
T
550
O
550
F
Arc CE and Arc EF are congruent because they are in the
same circle and have the same measure.
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Congruent Circles and Arcs
The two circles below are also called concentric circles, since
they share the same center, but have different radii lengths.
Concentric circles are similar, but not congruent.
T
R
O
C
E
R
S
U
Arc RS and Arc TU are similar since they have the same
measure.
But they are not congruent because they are arcs of circles
that are not congruent.
O
S
U
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15 Both Circle F and Circle G have a radius of 2.0 m.
Are Arc AB and Arc CD similar?
16 Both Circle F and Circle G have a radius of 2.0 m. Are
Arc AB and Arc CD congruent?
Yes
Yes
No
No
C
B
750
A
F
G
750
C
B
D
A
750
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17 AB # DC
F
G
750
D
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18 LP ≅ MN
True
True
False
False
A
M
1800
L
B
700
400
850
C
P
N
D
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19 Circle P has a radius of 3 and AB has a measureof
90°. What is the length of AB?
A
3√2
B
3√3
C
6
D
9
A
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20 Two concentric circles always have congruent
radii.
True
False
P
B
Note that you need to use the indicated degree
measures as the drawing is not to scale.
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21 If two circles have the same center, they are
congruent.
22 Tanny cuts a pie into 6 congruent pieces. What isthe
measure of the central angle of each piece?
True
False
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#
B
O
Arc Length & Radians
Before we extend our thinking of
central angles and arc measures to
arc lengths, it's worth reflecting on
the number π, which will be central
to our work.
This number was a devastating
discovery to Greek mathematicians.
A
In fact, the reason that The
Elements was written without relying
on numbers, was because numbers
were considered unreliable to the
Greeks after π was discovered.
Return to the
table of
contents
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#
B
O
A
Until then, Pythagoras, and his followers
believed that "All was Number."
But when they sought to find the number
that is ratio of the circumference to the
diameter of a circle, they found that
there wasn't one.
The closer they looked, the more
impossible it became to find a number
solution to the simple expression of
C/d: the circumference divided by the
diameter of a circle.
B
Until then, Pythagoras, and his
followers believed that "All was
Number."
O
But when they sought to find the
number that is ratio of the
http://bobchoat.files.wordpress.com/2013/06/pi-day004.jpg
A
circumference to the diameter of a
circle, they found that there wasn't
one.
The closer they looked, the more
impossible it became to find a
number that solves the simple
equation of C/d.
http://bobchoat.files.wordpress.com/2013/06/pi-day004.jpg
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#
#
π is an example of an irrational number. A
number that is not the ratio of two integers.
So, no matter how far you take it, it keeps
B going without settling down.
O
We are comfortable with irrational numbers
now, but the Greeks weren't.
A
Now we know that there are many more
irrational numbers than rational numbers.
In mathematics, it's best to just leave
answers with the symbol π.
B In science, engineering and other fields
which need a rational answer, and where π
shows up a lot, the value of π is just
estimated with the number of digits
necessary for the problem.
O
A
For most problems, 3.14 is close enough.
Rational numbers are like islands in a sea of
irrational numbers.
For others, you might use 3.14159...but you
will rarely need more than that.
But, we are more familiar with those islands
as that's where we grew up.
For this course, just leave your answers with
the number π as part of your answer.
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Arc Lengths
Arc Lengths
The relationship between the
circumference of a circle to its diameter is
B
O
A
C = πd
Since d = 2r, this is usually expressed as
A
O
B
700
But, if we are also told that the radius of
the circle is 20 cm, we can determine the
length of Arc AB, also denoted as AB.
C = 2πr
We know that a full trip around a circle is
equal to 3600, so if we know the angle of
an arc and the radius, we can determine
the length of the arc.
That's how far you'd have to travel along
that arc to get from Point A to Point B.
C
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Arc Lengths
Arc Lengths
We could do this by first figuring out the
circumference using
A
O
70
0
20
B
cm
Alternatively, we could just set this up as a
ratio and solve it in one step.
A
C = 2πr = 2π(20 cm) = 40π cm
Then figuring what percentage of the
circumference Arc AB is by the ratio of
O
(AB)/C = 700/3600 = 0.1944
C
Since 3600 is the entire circumference.
In the figure off to the left, we know that
the measure of Arc AB is equal to that of
the Central Angle...700.
C
70
0
20
B
cm
Arc length = Central angle
Circumference
360
AB = 70
2πr 360
70 (2πr)
AB =
360
So, AB is
70 (2π)(20)
AB =
360
(0.1944)(40π cm) = 7.8π cm (about 24 cm)
AB = 7.8π cm
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Example
Arc Lengths
For any arc, you can find its length by
multiplying the circumference of the circle
(2πr) by the angle of arc divided by 360.
A
B
O
#
In Circle A, the central angle is 600 and the radius is 8 cm.
Find the length of Arc CT
In this case, the measure of the arc is θ,
since it is equal to the central angle.
C
Then,
C
AB
2πr
=
AB
=
8 cm
θ
3600
A
600
θ
2πr
3600
T
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23 In circle C where AB is a diameter, find the length
of Arc BD. The length of diameter AB is 15 cm.
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24 In circle C where AB is a diameter, find the length of
Arc DAB. The length of AB is 15 cm.
B
B
1350
D
1350
D
C
15 cm
C
15 cm
A
A
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25 In circle C where AB is a diameter, find the length
of Arc ADB. The length of AB is 15 cm.
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26 In circle C can it be assumed that AB is a diameter?
Yes
No
B
B
1350
D
C
15 cm
1350
D
C
A
A
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28 Find the circumference of circle T.
27 Find the length of AB
A
B
450
T
3 cm
750
C
6.82 cm
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Radians
29 In circle T, Lines WY & XZ are diameters and have lengths of
6. The m∠XTY = 1400, what is YZ?
A
A
B
C
D
2 π
3
6π
4 π
3
4π
C
Another way of measuring
angles, as an alternative to
degrees, is radians.
Where degrees are arbitrary
(Why are there 360 degrees in
circle?), radians are very
natural.
O
Radians are the ratio of an arc
length divided by the radius of
the arc.
Draw a Picture.
Hint:
click to reveal
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Radians
A
C
So, the radian measure of angle
AOC is just the length of Arc AC
divided by the length of radius
(AO or CO).
The circumference of a circle is
given by C = 2πr.
O
Radians
So, the radian measure of a full
trip around a circle is 2πr/r = 2π.
A
C
O
There are no units for radians,
since the lengths cancel out, but
you can write rads or radians just
to indicate what you are doing.
Since there are no units, these
angle measures are much easier
to use when you study
trigonometry, physics and
calculus.
All scientific calculators allow you
to use degrees or radians. Just
make sure it is set to the correct
one when you are entering angle
measurements.
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Radians
A
Radians
Since a full trip around a circle
is 3600, and is also 2π radians,
they must be equal.
A
Alternatively, you could write a
proportion and use the crossproduct property to find your
missing measurement.
So, to convert from one to the
other, just multiply by the
appropriate conversion factor.
C
2π
=
3600
O
3600
2π
=1
C
O
degrees radians
= 2π
3600
Since 2π = 3600, each of these
fractions is just equal to 1.
Both methods will be shown in
the next example. Use
whichever method is easier for
you.
Multiplying anything by them
doesn't change it's value, since
multiplying by 1 has no effect.
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Radians
A
Radians
2π
=
3600
3600
2π
=1
A
For instance, m# AOC = 1250.
C
It is also equal to
O
m# AOC = (1250) 2π 0
360
C
If you are using the proportion and
the cross-product property, you
could set up & solve the problem
with the works shown below.
degrees radians
= 2π
3600
O
125° = x
2π
360°
m# AOC = 0.69 π radians
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30 How many radians is 180 degrees?
(Round π to 3.14 in your answer.)
250π = 360x
x = 25π = 0.69π radians
36
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31 How many radians is 90 degrees?
(Round π to 3.14 in your answer.)
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32 How many radians is 140°?
(Round π to 3.14 in your answer.)
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33 How many degrees is π radians?
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34 How many degrees is 2π radians?
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35 How many degrees is 1.0 radian?
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36 How many degrees is 1.6 radians?
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37 Circle F has a radius of 3.
What is the arc length for #AFB?
A π/2
A
B π
C 2π
D 3π/4
E
120°
F
B
45°
30°
C
D
PARCC Released Question - Response Format
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38 Circle F has a radius of 3.
39 Circle F has a radius of 3.
What is the arc length for #BFC?
What is the arc length for #CFD?
A π/2
A π/2
A
B π
C 2π
D 3π/4
A
B π
C 2π
120°
E
F
45°
30°
D 3π/4
B
120°
E
F
30°
C
D
C
D
PARCC Released Question - Response Format
PARCC Released Question - Response Format
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Question 3/7
40 Circle F has a radius of 3.
Topic: Angles, Arcs and Arc Lengths
What is the arc length for #AFE?
A
E
A π/2
F
120°
45°
30°
A
B π
B
45°
B
C
D
C 2π
D 3π/4
E
120°
F
B
45°
30°
C
D
PARCC Released Question - Response Format
PARCC Released Question
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Click on the link below and complete the
labs before the Chords lesson.
Chords, Inscribed
Angles & Triangles
Return to the
table of
contents
Lab - Properties of Chords
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The Arc of the Chord
Definition: When a chord and a minor arc have the same
endpoints, we call the arc The Arc of the Chord.
Chord Bisector Theorem
If a diameter or radius of a circle is perpendicular to a chord, then
the diameter or radius bisects the chord and its arc.
P
Q
C
PQ is the arc of PQ
O
A
CS is a diameter of the circle
and is perpendicular to chord AE
S
Therefore, XE ≅ XA & AS ≅ ES
O
X
Recall the definition of a chord:
a line segment with endpoints
on the circle.
E
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Proof of Chord Bisector Theorem
C
A
O
X
Proof of Chord Bisector Theorem
Given: CS is a diameter which is
perpendicular to AE
C
A
O
Prove: CS bisects AE and its
minor arc AE
S
X
E
As a first step, let's draw radii
from the Center O to the points A
and E to construct ΔAOE.
S
E
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41 Fill in the statement for step #2.
A ∠OXA and ∠OXE are complementary
B ∠OXA and ∠OXE are right angles
C
C
A
O
Now, we can use the triangles
which have been formed in our
proof .
E
S
X
S
E
Statement
1
X
A
O
C OX # OX
D OA # CO
E ΔOXA # ΔOXE
Reason
?
Right angles are formed by
perpendicular lines
4
?
Reflexive Property of #
5
?
Hypotenuse - Leg Theorem
3 OE # OA
?
6 XE # XA and ∠XOE # ∠XOA
7 SE # SA
8 CS bisects AE and Minor Arc AE
B
[This object is a pull tab]
CS is a diameter which is perpendicular
Given
to AE
2
Answer
Proof of Chord Bisector Theorem
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?
Arcs intercepted by # central
∠s of a circle are #
?
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42 Fill in the reason for step #3.
43 Fill in the statement for step #4.
A
A ∠OXA and ∠OXE are complementary
B ∠OXA and ∠OXE are right angles
S
C OX # OX
D OA # CO
E ΔOXA # ΔOXE
C
B Definition of supplementary
O
C Corresponding parts of # Δs are #
D Side-Angle-Side Theorem
Answer
X
E Definition of bisector
E
Reason
4
?
Reflexive Property of #
5
?
Hypotenuse - Leg Theorem
3 OE # OA
8 CS bisects AE and Minor Arc AE
?
Right angles are formed by
perpendicular lines
4
?
Reflexive Property of #
5
?
Hypotenuse - Leg Theorem
?
6 XE # XA and ∠XOE # ∠XOA
Arcs intercepted by # central
∠s of a circle are #
7 SE # SA
Reason
3 OE # OA
?
Arcs intercepted by # central
∠s of a circle are #
8 CS bisects AE and Minor Arc AE
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45 Fill in the reason for step #6.
A Radii of circles are #
C
O
C OX # OX
D OA # CO
E ΔOXA # ΔOXE
X
S
E
Statement
[This object is a pull tab]
Reason
CS is a diameter which is perpendicular
1
Given
to AE
Right angles are formed by
perpendicular lines
2
?
?
Right angles are formed by
perpendicular lines
3 OE # OA
?
4
?
Reflexive Property of #
4
?
Reflexive Property of #
5
?
Hypotenuse - Leg Theorem
5
?
Hypotenuse - Leg Theorem
6 XE # XA and ∠XOE # ∠XOA
?
6 XE # XA and ∠XOE # ∠XOA
Arcs intercepted by # central
∠s of a circle are #
7 SE # SA
8 CS bisects AE and Minor Arc AE
8 CS bisects AE and Minor Arc AE
Slide 101 / 259
Proof of Chord Bisector Theorem
Given: CS is a diameter which is
perpendicular to AE
S
Prove: CS bisects AE and its
minor arc AE
O
X
E Definition of bisector
E
1
Reason
?
3 OE # OA
Right angles are formed by
perpendicular lines
4
?
Reflexive Property of #
5
?
Hypotenuse - Leg Theorem
7 SE # SA
8 CS bisects AE and Minor Arc AE
A
O
X
E
S
E
Statement
?
6 XE # XA and ∠XOE # ∠XOA
C
[This object is a pull tab]
CS is a diameter which is perpendicular
Given
to AE
2
Answer
A
C
C Corresponding parts of # Δs are #
D Side-Angle-Side Theorem
Statement
?
Slide 102 / 259
46 Fill in the reason for step #8.
B Definition of supplementary
?
Arcs intercepted by # central
∠s of a circle are #
7 SE # SA
?
A Radii of circles are #
S
C
[This object is a pull tab]
Reason
3 OE # OA
X
E Definition of bisector
E
CS is a diameter which is perpendicular
Given
to AE
?
A
O
C Corresponding parts of # Δs are #
D Side-Angle-Side Theorem
E
2
C
B Definition of supplementary
A
Answer
A ∠OXA and ∠OXE are complementary
B ∠OXA and ∠OXE are right angles
Statement
?
Slide 100 / 259
44 Fill in the statement for step #5.
1
?
7 SE # SA
?
C
[This object is a pull tab]
CS is a diameter which is perpendicular
Given
to AE
2
?
6 XE # XA and ∠XOE # ∠XOA
Statement
1
Right angles are formed by
perpendicular lines
S
E
[This object is a pull tab]
?
2
X
A
CS is a diameter which is perpendicular
1
Given
to AE
A
O
Answer
Statement
C
Answer
A Radii of circles are #
?
Arcs intercepted by # central
∠s of a circle are #
?
1
Reason
CS is a diameter which is perpendicular
Given
to AE
2 ∠OXA and ∠OXE are right angles
Right angles are formed by
perpendicular lines
3 OE # OA
Radii of a circle are #
4 OX # OX
Reflexive Property of #
5 ΔOXA # ΔOXE
Hypotenuse - Leg Theorem
6 XE # XA and ∠XOE # ∠XOA
Corresponding parts of #
Δs are #
7 SE # SA
Arcs intercepted by #
central ∠s of a circle are #
8 CS bisects AE and Minor Arc AE
Definition of bisector
Slide 103 / 259
Slide 104 / 259
Converse of Chord Bisector Theorem
In a circle, if one chord is a perpendicular bisector of another chord,
then the first chord is a diameter.
C
A
O
X
S
The chord CS is the perpendicular
bisector of the chord AE.
Proof of
Converse of Chord Bisector Theorem
In a circle, if one chord is a perpendicular bisector of another chord,
then the first chord is a diameter.
C
Therefore, CS is a diameter of the
circle and passes through the center
of the circle.
A
O
X
E
S
E
Slide 105 / 259
B
Proof of Arcs and Chords Theorem
In the same circle, or in congruent circles, two minor arcs are
congruent if and only if their corresponding chords are congruent.
This follows from the fact that the
measure of an arc is equal to that of
the central angle which intercepts it.
AB ≅ CD iff AB ≅ CD
*iff stands for "if and only if"
C
Since all the radii, BO, AO, CO and
DO are congruent and the sides BA
and CD are congruent, the triangles
ABO and DOC are congruent by
Side-Side-Side.
C
B
O
O
D
A
That means that the central angles
are congruent which means that the
arcs are congruent since arcs
intercepted by congruent central
angles are congruent.
D
A
Slide 107 / 259
Slide 108 / 259
BISECTING ARCS
If XY ≅ YZ, then point Y and any line
segment, or ray, that contains Y,
bisects XYZ
EXAMPLE
Y
CD, DE, and CE
(9x)0
This just follows from the
definition of a bisector as
dividing something into two
pieces of equal measure.
A
D
E
Z
Find:
C
B
X
C
Construct ΔOAX and ΔOEX and by
proving them congruent, you show
that OA and OE are radii, which
means that chord CS passes
through the center, and is a
diameter.
Slide 106 / 259
Arcs and Chords Theorem
In the same circle, or in congruent circles, two minor arcs are
congruent if and only if their corresponding chords are congruent.
This proof is very much like that of
the original theorem of which this is
the converse.
(80 - x)0
Slide 109 / 259
Theorem
In the same, or congruent, circles, two chords are
congruent if and only if they are equidistant from the center.
Slide 110 / 259
Theorem
In the same, or congruent, circles, two chords are
congruent if and only if they are equidistant from the center.
AB ≅ CD iff EF ≅ EG
AB ≅ CD iff EF ≅ EG
C
C
G
D
E
B
G
Proving this just requires creating
Δ AEB and ΔCED and proving
that they are congruent, which
means that their altitudes are
equal, which is their distance
from the center of the circle.
D
Here Δ AEB and ΔCED and their
congruent sides are shown.
E
B
F
F
A
A
Slide 111 / 259
Example
Given circle C, QR = ST = 16.
Find CU.
Slide 112 / 259
47 In circle R, AB ≅ CD and mAB = 108°. Find mCD.
A
R
U
Q
R
T
V
S
C
1080
2x
C
5x - 9
B
D
Slide 113 / 259
48 Given circle C below, the length of BD is:
A
5
B
C
D
10
15
20
D
A
B
10
C
F
Slide 114 / 259
49 Given: circle P, PV = PW, QR = 2x + 6, and
ST = 3x - 1. Find the length of QR.
A
1
B
7
C
20
D
8
R
V
.
Q
S
P
W
T
Slide 115 / 259
Slide 116 / 259
Inscribed Angles
50 AH is a diameter of the circle.
A
A
True
False
3
M
3
Inscribed angles are angles
whose vertices lie on the circle
and whose sides are chords of
the circle.
5
S
Angle ABC is an inscribed
angle and Arc AC is its
intercepted arc.
T
H
B
C
Slide 117 / 259
Slide 118 / 259
Inscribed Angle Theorem
Inscribed Angles
A
The measure of an inscribed angle is equal to half the
measure of the intercepted arc or central angle
The minor arc that lies in
the interior of the inscribed
angle and has endpoints
which are vertices of the
angle is called the
intercepted arc.
O
A
To prove this, we need to
find the measure of ∠ABC
relative to the measure of
the intercepted Arc AC
Arc AC is the intercepted
arc of inscribed angle
ABC.
B
B
C
C
Lab - Inscribed Angles
Slide 119 / 259
Slide 120 / 259
Inscribed Angle Theorem
Inscribed Angle Theorem
To prove the theorem, we
first draw the diameter BD
which creates ΔAOB and
ΔCOB
A
A
The measure of∠AOC,
shown with green lines, is
the same as the measure
of the minor arc AC, shown
in blue.
D
O
O
B
B
C
C
Slide 121 / 259
Slide 122 / 259
52 What is true of all radii of the same circle?
51 What are OB, OA, OD and OC?
A
chords
C
B
radii
D nothing special
diameters
A
the are perpendicular
C
B
they are parallel
D not special
they are of equal length
A
A
D
D
O
O
B
B
C
C
Slide 123 / 259
Slide 124 / 259
54 Which is true about base angles of isosceles Δs?
53 What types of triangles are ∆BOA and ∆BOC?
A scalene
B
A they are equal
C right
isosceles
B
D not special
A
they are double
C
they are half
D
nothing special
A
B
D
D
O
O
B
Why?
C
B
C
Slide 125 / 259
Inscribed Angle Theorem
A
Equal angles are
marked by x and y.
Slide 126 / 259
55 What types of angle are angles a and b of ∆ BOA and
∆BOC?
A
interior
C
B
exterior
D not special
A
x
D
x
D
O
B
right
x
y
O
y
B
C
x
y
a
b
y
C
Slide 127 / 259
Slide 128 / 259
57 To what is the measure of the sum of a + b equal?
56 Which of the below is true?
A
a = 2x
C
a=b
A
a + b = arc AD
C
a + b = arc AC
B
b = 2y
D
x=y
B
a + b = arc DC
D
a+b=x+y
A
A
x
x
D
a
O
B
D
O
b
x
y
x
y
B
a
b
y
y
C
C
Slide 129 / 259
Slide 130 / 259
Inscribed Angle Theorem
A
Also, ∠a+∠ b = 2x+2y = 2(x+y)
D
B
x
y
a
And x+y = m∠ABC
So, 2(m∠ABC) = m∠AOC
b
And m∠ABC = 1/2(m∠AOC)
y
C
The measure of the inscribed
angle is equal to half the
measure of the central
angle and, therefore, half the
measure of the intercepted arc.
x
B
C
This was proven if the
center is within the angle,
but is true for all cases.
Inscribed Angle Theorem
C
A
It's not essential to go
through the proof of the
other two cases, but if you
have time, it's good
practice.
x
First, if the
inscribed angle does not
include the center.
B
Also, it is equal to half the
measure of the central
angle intercepting that arc.
Slide 132 / 259
Inscribed Angle Theorem
C
2x
2x
Slide 131 / 259
A
The measure of an
inscribed angle, such as
∠ABC, is equal to half the
measure of the intercepted
arc, which is AC.
A
So, ∠a+∠b = ∠AOC
x
O
Inscribed Angle Theorem
y
ab
xy
B
O
Since these triangles
overlap so much one side
of the inscribed angle, AB,
and its associated
isosceles triangle and
external angle are shown in
D orange.
The other leg and its
associated isosceles
triangle and external angle
are shown in dark blue.
The proof then follows the
same form.
Slide 133 / 259
Slide 134 / 259
Inscribed Angle Theorem
Inscribed Angle Theorem
C
A
x
y
D
ab
O
xy
m∠ABC = x-y
m∠AOC = a-b
m∠AOC = 2x-2y = 2(x-y)
O
So,
B
m∠ABC = 1/2(m∠AOC)
C
The inscribed angle equals
half the measure of the
intercepted arc
A
Slide 135 / 259
Slide 136 / 259
Example
Inscribed Angle Theorem
m∠ABC = x
B
Find m∠T, mTQ and mQR.
a = 2x
x
m∠AOC = a = 2x
R
Q
500
So, m∠ABC = 1/2(m∠AOC)
O
a
x
The last case is if the a leg
of the inscribed angle
passes through the center.
B
a = 2x
b = 2y
The inscribed angle equals
half the measure of the
intercepted arc
C
480
P
S
T
A
Slide 137 / 259
Slide 138 / 259
Theorem
Theorem
If two inscribed angles of a circle intercept the same arc,
then the angles are congruent.
A
B
D
This follows directly from our
prior proofs showing that the
measure of an inscribed angle is
equal to half that of the
intercepted arc.
By the transitive property of
equality, if two inscribed angles
intercept the same are they must
be equal.
In this case, m∠ADB = m∠ACB
since the both intercept Arc AB.
C
∠ADB ≅ ∠ACB
In a circle, parallel chords intercept congruent arcs.
C
D
Note that this does NOT mean
that Arc CD is equal to Arc AB.
O
A
In circle O, if Chord AB is
parallel to Chord CD then Arc
CA is equal to Arc DB.
B
Slide 139 / 259
Slide 140 / 259
58 Given the figure below, which pairs of angles are
congruent?
A
∠R ≅ ∠S
∠U ≅ ∠T
B
∠R ≅ ∠U
∠S ≅ ∠T
C
∠R ≅ ∠T
∠U ≅ ∠S
D
∠R ≅ ∠T
∠R ≅ ∠S
59 Find m∠Y
R
X
S
Y
U
P
T
Z
Slide 141 / 259
60 In a circle, two parallel chords on opposite sides of the
center have arcs which measure 1000 and 1200.
Slide 142 / 259
61 Given circle O, find the value of x.
Find the measure of one of the arcs included between
the chords.
x
B
A
300
C
D
O
AB || CD
Slide 143 / 259
Slide 144 / 259
Try This
62 Given circle O, find the value of x.
In this circle:
mPQ = 52°
mQS = 112°
mST = 88°
1000
Find
m∠1, m∠2,
m∠3 & m∠4
Q
B
A
350
P
O
C
2
1
3
D
x
4
AB || CD
S
T
Slide 145 / 259
Slide 146 / 259
Inscribed Triangles
Corollary to Inscribed Angle Theorem
A triangle is inscribed if all its vertices lie on a circle.
If a right triangle is inscribed in a circle, then the
hypotenuse is a diameter of the circle.
.
.
inscribed
triangle
.
AG is a diameter of the circle if
ΔALG is a right Δ, and # ALG
is its right angle, follows
directly from the inscribed
angle theorem.
A
x
.
Since # ALG has a measure of
900, it must intercept an arc
whose measure is 1800, this is
half the circle, so AG must be
a diameter
L
G
m# ALG = 90°
Slide 147 / 259
Slide 148 / 259
63 In the diagram, ∠ADC is a central angle and
m∠ADC = 600. What is m∠ABC?
A 150
A
B 300
B
C 600
.
D
D 1200
C
64 What is the value of x?
A
5
B
10
C
13
D
15
Slide 149 / 259
65 If m #CBD = 440, find m#BAC.
The figure shows ΔABC inscribed in circle D.
B
C
D
A
PARCC Released Question - Response Format
E
(12x + 40)0
F
(8x + 10)0
G
Slide 150 / 259
Question 1/25
Topic: Inscribed Angles
Slide 151 / 259
Slide 152 / 259
66 Find the value of x.
67 The length of CD is _________ because....
C
10
B
C
A 10
B less than 10
D
P
C greater than 10
10
B
P
not to scale
not to scale
The figure shows a circle with center P, a diameter BD, and
inscribed ΔBCD. PC = 10. Let m# CBD = x0 and m# BCD = (x+54)0.
The figure shows a circle with center P, a diameter BD, and
inscribed ΔBCD. PC = 10. Let m# CBD = x0 and m# BCD = (x+54)0.
PARCC Released Question - Response Format
PARCC Released Question - Response Format
Slide 153 / 259
Slide 154 / 259
Question 17/25
68 The length of CD is ... because ________.
Topic: Inscribed Angles
C
A ΔCPD is equilateral
B m#CPD < 600
C m#CPD > 600
D
10
B
P
D
not to scale
The figure shows a circle with center P, a diameter BD, and
inscribed ΔBCD. PC = 10. Let m# CBD = x0 and m# BCD = (x+54)0.
PARCC Released Question - Response Format
a. 10
d. ∆CPD is equilateral
b. less than 10
e. m∠CPD < 60°
c. greater than 10
f. m∠CPD > 60°
Slide 155 / 259
Point B is the center of a circle, and AC is a diameter of the
circle. Point D is a point on the circle different from A and C.
69 The statement AD > CD is:
Slide 156 / 259
Point B is the center of a circle, and AC is a diameter of the
circle. Point D is a point on the circle different from A and C.
70 The statement m#CBD = 1/2 (m#CAD) is:
A Always True
A Always True
B Sometimes True
B Sometimes True
C Never True
C Never True
PARCC Released Question - Response Format
PARCC Released Question - Response Format
Slide 157 / 259
Slide 158 / 259
Point B is the center of a circle, and AC is a diameter of the
circle. Point D is a point on the circle different from A and C.
Point B is the center of a circle, and AC is a diameter of the
circle. Point D is a point on the circle different from A and C.
72 The statement that m#ABD = 2(m#CBD) is:
71 The statement that m#CBD = 900 is:
A Always True
A Always True
B Sometimes True
B Sometimes True
C Never True
C Never True
PARCC Released Question - Response Format
PARCC Released Question - Response Format
Slide 159 / 259
Slide 160 / 259
Point B is the center of a circle, and AC is a diameter of the
circle. Point D is a point on the circle different from A and C.
Question 22/25
Topic: Inscribed Angles
73 If m#BDA = 200, what is m#CBD)?
A 200
B 400
C 700
D 1400
PARCC Released Question - Response Format
Slide 161 / 259
Slide 162 / 259
Tangents
C
Tangents & Secants
I
B
Return to the
table of
contents
H
A
G
D
E
F
A tangent is a line, ray or segment
which touches a circle at just one
point.
All three types of tangent lines are
shown on this drawing.
The point where the line touches the
circle is called the "point of
tangency."
In this case, those points are B, E
and H.
D
Slide 163 / 259
Slide 164 / 259
NOT Tangents
Secants
Note that for a ray or
segment to be a tangent, it
must not touch the circle in
more than one point even if it
were extended.
C
The segment and ray shown
to the left are NOT tangents
because if they were
extended they would touch
the circle in more than one
point.
A
B
Slide 165 / 259
I
A secant is a line, ray or segment
which touches a circle at two points.
All three types of secant lines are
shown on this drawing.
A
D
G
F
Slide 166 / 259
Intersections of Circles
Intersections of Circles
Coplanar circles can intersect at zero, one, or two points.
Below are shown three ways in which a pair of circles can have
no points of contact.
Circles which are sideby-side may not intersect.
C
Tangent Circles intersect at one point.
The two types of tangent circles are shown below.
Circles within one another may not
intersect.
.
Remember that:
Circles within one
another and have a
common center are
"concentric."
Slide 167 / 259
Intersections of Circles
These circles intersect at two points.
Can two distinct circles intersect at three points?
Slide 168 / 259
Common Tangents
Two circles can have between zero and four common tangents.
Two completely separate circles have four common tangents.
Slide 169 / 259
Slide 170 / 259
Common Tangents
Common Tangents
Two circles can have between zero and four common tangents.
Two circles can have between zero and four common tangents.
Two externally tangent circles have three common tangents.
Two overlapping circles have two common tangents.
Slide 171 / 259
Common Tangents
Slide 172 / 259
74 How many common tangent lines do the circles have?
Two circles can have between zero and four common tangents.
Two non-tangent circles within one another have no common
tangents.
Slide 173 / 259
75 How many common tangent lines do the circles have?
Slide 174 / 259
76 How many common tangent lines do the circles have?
Slide 175 / 259
Slide 176 / 259
77 How many common tangent lines do the circles have?
78 Which term best describes Line BE?
A
B
C
D
E
F
G
H
Radius
Diameter
Chord
Secant
Tangent
Common Tangent
Point of Tangency
Center
B
A
D
F
C
G
E
Slide 177 / 259
Slide 178 / 259
79 Which term best describes Line DE?
80 Which term best describes Line AB?
A
B
C
D
E
F
G
H
A
B
C
D
E
F
G
H
Radius
Diameter
Chord
Secant
Tangent
Common Tangent
Point of Tangency
Center
B
Radius
Diameter
Chord
Secant
Tangent
Common Tangent
Point of Tangency
Center
A
B
A
D
D
F
C
F
C
G
E
G
E
Slide 179 / 259
Slide 180 / 259
81 Which term best describes Points D and G?
82 Which term best describes Line BF?
A
B
C
D
E
F
G
H
A
B
C
D
E
F
G
H
Radius
Diameter
Chord
Secant
Tangent
Common Tangent
Point of Tangency
Center
B
Radius
Diameter
Chord
Secant
Tangent
Common Tangent
Point of Tangency
Center
A
B
A
D
D
F
C
F
C
E
G
E
G
Slide 181 / 259
Slide 182 / 259
83 Which term best describes Line AC?
84 Which term best describes Point F?
A
B
C
D
E
F
G
H
A
B
C
D
E
F
G
H
Radius
Diameter
Chord
Secant
Tangent
Common Tangent
Point of Tangency
Center
B
Radius
Diameter
Chord
Secant
Tangent
Common Tangent
Point of Tangency
Center
A
B
A
D
D
F
C
G
E
Slide 184 / 259
Tangents and Radii are Perpendicular
85 Which term best describes Segment BA?
Radius
Diameter
Chord
Secant
Tangent
Common Tangent
Point of Tangency
Center
G
E
Slide 183 / 259
A
B
C
D
E
F
G
H
F
C
A tangent and radius that intersect on a circle are perpendicular.
This is an essential result for mathematics and physics. It appears to
be true, but let's prove it.
B
A
O
D
F
C
G
E
A
B
C
Slide 185 / 259
Slide 186 / 259
Proof that Tangents and Radii are
Perpendicular
Proof that Tangents and Radii are
Perpendicular
Let's use an indirect proof.
If OD is a leg of the right triangle
and OB is the hypotenuse, then
OB must be longer than OD.
We'll make an assumption and
see if it leads to a contradiction.
O
A
B
D
But, OD extends from the center of
the circle to beyond the circle.
Let's assume that another point on
AC is where a line from Point O is
perpendicular to the AC.
Let's name that point D, so the line
perpendicular to AC is OD.
Then ∆ODB must be a right
triangle with OD and BD being the
legs.
C
Then OB must be the hypotenuse.
And OB only extends from the
center of the circle to the
circumference of the circle.
O
So, OB must be shorter than OD.
A
B
D
C
This is a contradiction, which
proves that our original
assumption was incorrect.
Slide 187 / 259
Slide 188 / 259
Proof that Tangents and Radii are
Perpendicular
Using an Intersecting Tangents
& Radius to Solve Problems
Our assumption was that OB
was not perpendicular to AC.
Whenever you are given, or can
draw, a circle and a tangent, you
can construct a radius to the
point of tangency.
If that is false, then it must be
that OB is perpendicular to AC,
which is what set out to prove.
O
The tangent and radius will form
a right angle.
O
A
This is often very helpful and
what is needed to solve a
problem.
B
C
A
B
C
Lab - Tangent Lines
Slide 189 / 259
Slide 190 / 259
Using an Intersecting Tangent
& Radius to Solve Problems
Velocity
Gravity
Sometimes, that takes the form
of creating a right triangle, with
all the information that is
provided.
Using an Intersecting Tangent
& Radius to Solve Problems
This property of a tangent and
intersecting radius shows up
often in physics.
Or, the maximum torque that can
be applied on a wheel by a road
when braking, or by a tire on a
road while accelerating.
For instance, it explains why the
force of gravity pulling the moon
into its circular orbit is
perpendicular to the direction of
its orbital velocity.
The force acts along the radius
and the motion is tangent to the
circular orbit.
Slide 191 / 259
Using an Intersecting Tangent
& Radius to Solve Problems
Slide 192 / 259
Theorem
Tangent segments from a common external point are congruent.
∠R and ∠T are right angles.
R
Or, the transfer of force due to a
bicycle chain.
PA is congruent to itself by
the Reflexive Property.
P
A
T
PR and PT are congruent
because they are radii.
ΔPRA and ΔPTA are
congruent due to the
Hypotenuse-Leg Theorem.
Therefore, AR and AT are
congruent since they are
corresponding parts of
congruent triangles.
Slide 193 / 259
Slide 194 / 259
Example
86 AB is a radius of circle A. Is BC tangent to circle A?
Given: RS is tangent to circle C at S and RT is tangent to circle C
at T. Find x.
Yes
No
S
B
60
25
C
67
A
28
C
R
3x + 4
T
Slide 195 / 259
Slide 196 / 259
87 S is a point of tangency. Find the radius r of
circle T.
A
A
31.7
B
60
T
C
14
r
D
3.5
S
25
r
C
48 cm
89 AB, BC, and CA are tangents to circle O. AD = 5,
AC = 8, and BE = 4. Find the perimeter of triangle
ABC.
B
R
B
3x - 8
Slide 198 / 259
Angles Intercepted by
Tangents and Secants
Tangents and secants can form other angle
relationships in circle. Recall the measure of an
inscribed angle is 1/2 its intercepted arc.
This can be extended to any angle that has its vertex
on the circle.
F
This includes angles formed by two secants, a secant
and a tangent, a tangent and a chord, and two
tangents.
E
O
D
D
36 cm
Slide 197 / 259
C
88 In circle C, DA is tangent at A and DB is tangent at B.
Find x.
A
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Slide 200 / 259
Theorem: A Tangent and a Chord
Theorem
If a tangent and a chord intersect at a point on a circle, then the
measure of each angle formed is one half the measure of its
intercepted arc.
If a tangent and a secant, two tangents, or two secants
intersect outside a circle, then the measure of the angle
formed is half the difference of their intercepted arcs.
This is just a special
case of the rule that the
measure of an
inscribed angle is equal
to half the measure of
its intercepted arc.
M
A
This results in three rules for calculating angles, but they are
very similar to one another, so not hard to remember.
In this case,
m∠1 = 1 mRM
2
2 1
R
Especially, if you see one right after the other.
m∠2 = 1 mMAR
2
Slide 201 / 259
Slide 202 / 259
The Intercepted Arc of a
Tangent and a Secant
The Intercepted Arc of
Two Tangents
B
m∠1 =
A
P
1
(mBC - mAC)
2
Q
1
m∠2 =
1
(mPQM - mPM)
2
2
C
M
Slide 203 / 259
The Intercepted Arc of
Two Secants
Slide 204 / 259
The Intercepted Arc of Two Chords
If two chords intersect inside a circle, then the measure of each
angle is the average of the two intercepted arcs.
A
X
M
m∠3 =
W
1
(mXY - mWZ)
2
1
m∠1 =
1
1
(mAT + mMH)
2
3
Z
Y
H
T
This is equally true for each
pair of vertical angles.
The other pair of angles is
on the next slide.
Slide 205 / 259
Slide 206 / 259
Chords vs. Secants and/or Tangents
The Intercepted Arc of Two Chords
If two chords intersect inside a circle, then the measure of each
angle is the average of the two intercepted arcs.
A
A way to remember whether to add or subtract the two arcs that
the segments intersect is to visualize the segments rotating on
their intersection point to see which operation sign they are close
to resembling.
A
M
X
M
2
m∠2 =
2
2
1
(mMA + mHT)
2
W
2
3
H
H
Z
T
T
In the diagram above, if you
manipulate the 2 segments at the
intersection point, they make an
addition sign, so add the arcs
together before taking 1/2.
Slide 207 / 259
90 Find the value of x.
Y
In the diagram above, if you
manipulate the 2 rays at their
intersection point, they overlap
each other, making a subtraction
sign, so subtract the arcs before
taking 1/2.
Slide 208 / 259
91 Find the value of x.
1 3 00
D
x0
C
760
1780
x0
B
A
1 5 60
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92 Find the value of x.
Slide 210 / 259
93 Find the value of x.
C
780
H
E
x0
420
D
F
( 3 x - 02 )
( x + 06 )
3 40
Slide 211 / 259
Slide 212 / 259
95 Find m#1.
94 Find mAB.
B
2 6 00
1
650
A
Slide 213 / 259
96 Find the value of x.
0
12
x
2.
Slide 214 / 259
Example
To find the angle, you need the measure of both intercepted arcs.
First, find the measure of the minor arc mAB. Then we can
calculate the measure of the angle x0.
5
4 50
B
2470
x0
A
Slide 215 / 259
97 Find the value of x.
Slide 216 / 259
98 Find the value of x.
Students type their answers here
Students type their answers here
2 2 00
x0
x0
1 0 00
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Slide 218 / 259
99 Find the value of x
100 Find the value of x.
Students type their answers here
5 00
Students type their answers here
x0
1 2 00
( 5 x + 10 0 )
Slide 219 / 259
Slide 220 / 259
Released PARCC Exam Question
101 Find the value of x.
The following question from the released PARCC exam
uses what we just learned and combines it with what we
learned earlier to create a challenging question.
( 2 x - 300 )
3 00
Please try it on your own.
x
Then we'll go through the process we used to solve it.
Slide 221 / 259
Question 5/25
Slide 222 / 259
Topic: Tangents
Question 5/25
Topic: Tangents
This is a great problem and draws on a lot of what we've learned.
Take a shot at it in your groups.
Then we'll work on it step by step together by asking questions
that break the problem into pieces.
PB = BC = 6
Slide 223 / 259
102 What have we learned that will help solve this problem?
Slide 224 / 259
Question 5/25
Topic: Tangents
A A tangent and intersecting radius are perpendicular
B All radii of a circle are congruent
C Similar triangles have proportional corresponding parts
D All of the above
First, draw a radius from the center of each circle
to the point where the given tangent meets the
circle, points F and P.
PB = BC = 6
PB = BC = 6
Slide 225 / 259
Question 5/25
Topic: Tangents
Slide 226 / 259
103 What can you see from this drawing?
A ΔFDP and ΔEDM are right triangles
Now draw ΔFDP and ΔEDM.
B ΔFDP and ΔEDM are similar
C The corresponding sides of FDP and EDM are proportional
D All of the above
PB = BC = 6
PB = BC = 6
Slide 227 / 259
104 What are the lengths of FP and EM?
Slide 228 / 259
105 What is the ratio of proportionality, k, between ΔFDP and ΔEDM?
A FP = 6 and EM = 6
B FP = 3 and EM = 3
C FP = 6 and EM = 3
D They can't be found from this
PB = BC = 6
PB = BC = 6
6
3
6
3
3
x
Slide 229 / 259
Slide 230 / 259
106 Write an expression for the length of PD.
107 Write an expression for the length of MD.
A PD = 12
A MD = 12
B PD = 12 + x
B MD = 3 + x
C PD = 6 + x
D It can't be found from this
C PD = 6 + x
D It can't be found from this
PB = BC = 6
PB = BC = 6
6
6
3
6
3
3
3
6
x
Slide 231 / 259
3
3
x
Slide 232 / 259
108 Write a proportion relating MD and PD.
6 = 12 + x
A
3
3+x
B 3 = 12 + x
6
3+x
C 6 = 12 + x
3
3
D It can't be found from this
109 What is the length of MD?
PB = BC = 6
PB = BC = 6
6
6
3
6
3
6
3
110 Write an equation to find ED.
A ED = 3 + 9
2
3
3
x
x
Slide 233 / 259
2
3
Slide 234 / 259
111 What is the length of ED?
2
B 92 = ED2 - 32
C 92 = ED2 + 32
D It can't be found from this
PB = BC = 6
PB = BC = 6
6
3
3
6
9
3
3
x
Slide 235 / 259
Slide 236 / 259
Question 5/25
112 What would be a good way to check your answer.
Topic: Tangents
A Solve for FD and see if it is equal to ED
B Solve for PD and see if it is equal to ED
C Solve for FD and see if it is double ED
D Solve for FD and see if it is half ED
PB = BC = 6
6
3
6
3
3
x
Slide 237 / 259
Slide 238 / 259
Theorem
If two chords intersect inside a circle, then the products of the
measures of the segments of the chords are equal.
C
A
E
Return to the
table of
contents
B
D
Slide 239 / 259
113 Find the value of x.
114 Find the length of ML.
M
4
5
Slide 240 / 259
5
x+2
x
L
J
x
K
x+4
x+1
Answer
Segments & Circles
Slide 241 / 259
Slide 242 / 259
116 Find the value of x.
115 Find the length of JK.
x
M
L
J
x
16
K
x+2
9
18
x+4
x+1
Slide 243 / 259
118 Find the value of x.
117 Find the value of x.
A
-2
B
4
C
5
D
6
Slide 244 / 259
x
2
x
+
+
1
1 92 x
2
+
6
x
2x + 6
2
x
Slide 245 / 259
Secant Segment Theorem
If two secant segments are drawn to a circle from an external point,
then the product of the measures of one secant segment and its
external secant segment equals the product of the measures of the
other secant segment and its external secant segment.
B
A
E
C
D
Slide 246 / 259
119 Find the value of x.
9
6
x
5
Slide 247 / 259
Slide 248 / 259
120 Find the value of x.
121 Find the value of x.
3
5
x + 2
x + 1
4
x + 4
x - 1
x - 2
Slide 249 / 259
Tangent-Secant Theorem
Slide 250 / 259
122 Find the length of RS.
Q
If a tangent segment and a secant segment are drawn to a circle
from an external point, then the square of the measure of the
tangent segment is equal to the product of the measures of the
secant segment and its external secant segment.
16
R
x
T
8
S
A
C
E
D
Slide 251 / 259
123 Find the value of x.
Slide 252 / 259
124 Find the value of x.
24
3
1
x
12
x
Slide 253 / 259
Slide 254 / 259
Question 3/7
Topic: Angles, Arcs and Arc Lengths
A
Questions
from
Released
PARCC
Examination
E
F
120°
45°
30°
B
C
D
Return to the
table of
contents
Slide 255 / 259
Question 1/25
Topic: Inscribed Angles
Slide 256 / 259
Question 5/25
Slide 257 / 259
Question 17/25
Topic: Inscribed Angles
a. 10
d. ∆CPD is equilateral
b. less than 10
e. m∠CPD < 60°
c. greater than 10
f. m∠CPD > 60°
Topic: Tangents
Slide 258 / 259
Question 22/25
Topic: Inscribed Angles
Slide 259 / 259