Unit 10 -Circles
•This unit addresses circles.
•It includes central angles, arcs
(minor/major/semicircle), arc lengths,
sectors, areas of sectors, segments,
tangents to circles, circumscribed and
inscribed circles, and chords.
Standards
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SPI’s taught in Unit 10:
SPI 3108.1.1 Give precise mathematical descriptions or definitions of geometric shapes in the plane
and space.
SPI 3108.4.8 Solve problems involving area, circumference, area of a sector, and/or arc length of a
circle.
SPI 3108.4.13 Identify, analyze and/or use basic properties and theorems of circles to solve
problems (including those relating right triangles and circles).
CLE (Course Level Expectations) found in Unit 10:
CLE 3108.1.7 Use technologies appropriately to develop understanding of abstract mathematical
ideas, to facilitate problem solving, and to produce accurate and reliable models.
CLE3108.2.1 Establish the relationships between the real numbers and geometry; explore the
importance of irrational numbers to geometry.
CLE 3108.3.1 Use analytic geometry tools to explore geometric problems involving parallel and
perpendicular lines, circles, and special points of polygons.
CLE 3108.4.9 Develop the role of circles in geometry, including angle measurement, properties as a
geometric figure, and aspects relating to the coordinate plane.
CLE 3108.5.1 Analyze, interpret, employ and construct accurate statistical graphs.
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Standards
CFU (Checks for Understanding) applied to Unit 10:
3108.1.5 Use technology, hands-on activities, and manipulatives to develop the language and the
concepts of geometry, including specialized vocabulary (e.g. graphing calculators, interactive geometry
software such as Geometer’s Sketchpad and Cabri, algebra tiles, pattern blocks, tessellation tiles,
MIRAs, mirrors, spinners, geoboards, conic section models, volume demonstration kits, Polyhedrons,
measurement tools, compasses, PentaBlocks, pentominoes, cubes, tangrams).
3108.1.7 Recognize the capabilities and the limitations of calculators and computers in solving
problems.
3108.2.1 Analyze properties and aspects of pi (e.g. classical methods of approximating pi, irrational
numbers, Buffon’s needle, use of dynamic geometry software).
3108.2.2 Approximate pi from a table of values for the circumference and diameter of circles using
various methods (e.g. line of best fit).
3108.3.3 Find the equation of a circle given its center and radius and vice versa.
3108.4.13 Locate, describe, and draw a locus in a plane or space (e.g., fixed distance from a point on
a plane, fixed distance from a point in space, fixed distance from a line, equidistant from two points,
equidistant from two parallel lines, and equidistant from two intersecting lines).
3108.4.40 Find angle measures, intercepted arc measures, and segment lengths formed by radii,
chords, secants, and tangents intersecting inside and outside circles.
3108.4.41 Use inscribed and circumscribed polygons to solve problems concerning segment length
and angle measures.
3108.5.1 Determine the area of each sector and the degree measure of each intercepted arc in a pie
chart.
3108.5.2 Translate from one representation of data to another (e.g., bar graph to pie graph, pie graph
to bar graph, table to pie graph, pie graph to chart) accurately using the area of a sector.
Review of Circles in a Plane
• A circle is a set of points equally distant from a
a
center point
m
• A circle is named by it’s center point
• A radius is a segment that has one endpoint in
the center, and one on the circle.
• Congruent circles have congruent radii (or
diameters)
• A diameter is a segment that contains the center
of a circle and has both endpoints on the circle.
• A central angle is an angle whose vertex is the
center of the circle.
.
Find the Measure of the Central
Angle
• A study of 3600 people shows
that this is how most people
spend their time. The question
Other
is, what is the measure of each
15%
Sleep
central angle used to make
31%
Entertainment
these pie slices?
18%
• Sleep = 31% of 360=
Must Do
7%
Food
• .31x360 = 111.6
Work
9%
20%
• Food = 9% of 360=
• .09x360 = 32.4
• Work = 20% of 360=
How would I figure out how many people
• .20x360 = 72
are in each category?
• And so on…
Arcs
• An arc is a part of a circle
• One type of arc is a semicircle. A
semicircle is half of a circle.
• The measure of a semicircle is 180 degrees
• A minor arc is smaller than a semicircle
• The measure of a minor arc is the measure
of its corresponding central angle
• A major arc is greater than a semicircle
• The measure of a major arc is 360 minus
the measure of its related minor arc
Identifying Arcs
• Identify the following in
circle O
• Minor Arcs:
A
C
– AD, CE, AC, DE
• Semicircles:
– ACE, CED, EDA, DAC
• Major Arcs containing
point A:
– ACD, CEA, EDC, DAE
O
D
E
Arc Addition Postulate 7.1
• Adjacent Arcs are arcs of the same circle that have
exactly ONE POINT in common.
• The measure of the arc formed by two adjacent
arcs is the sum of the measures of the two arcs.
• Remembering that to measure an arc, you take the
measure of the corresponding central angle, then
the sum of the measures of two adjacent arcs is
really the sum of the measures of two adjacent
central angles.
• And just like adjacent angles share one side,
adjacent arcs share one point.
• Congruent arcs are arcs that have the same
measure AND are in the same circle or in
congruent circles
Find the Measure of the Arc
• BC
• 32
• BD
B
• 32 + 58 = 90
580
C
320
O
• ABC
• ABC is a semicircle so 1800
• AB
• 180 – 32 = 148
• ADB
• 180 + 32 = 212
A
D
Circumference and Arc Length
• Remember, circumference of a circle is
π x the diameter
• Theorem: The length of an arc of a circle is
calculated like this:
(measure of the arc) X π x D
360
• This is the same as saying the “fraction of
the whole” i.e. ¾ x the circumference of a
circle (which is π x D)
Find the Arc Length
• Find the length of xy
• Length of XY =
(mXY0/3600)x (π x D)
= 90/360 x (π x 16)
=.25 x (π x 16)
= 4π inches
X
O
900
Y
Find the length of the arc
•
•
•
•
•
•
Find the length of arc XPY
X
The radius here is 15 cm
= (mXPY0/3600)x (π x 2r)
P.
= (240/360) x (π x 2(15))
240
= 2/3 x 30 π
= 20π cm
Remember, Diameter = 2 Radii
O 15 CM
Y
0
Of course... There is a way to do it on the calculator… 
Assignment
• Page 654 9-27
• Page 655 30-35,37-43
• Worksheet 7-6
Unit 10 Quiz 1-round all
answers to the nearest 10th
• Using Circle O, find the following information:
1. Measure of arc AD in degrees
2. Measure of arc AB in degrees
B
3. Measure of arc ABD in degrees
4. Measure of arc ADC in degrees
5. Circumference of Circle O in pi
6. Length of arc BC in pi
7. Length of arc CD in pi
8. Length of arc DA in pi
9. Length of arc BDA in pi
10.If you rolled this circle 25 times, how far would
you roll it? In inches (rounded to the nearest
10th)
580
C
320
O
A
D
Areas of Circles and Sectors
• Imagine taking a circle and cutting it into
four quarters.
• Cut each quarter into four wedge segments
• Tape the wedges together to form a rough
rectangle
• Note that the areas are still the same
• The base of this figure (b) is formed by the
sum of the arcs of the circle  or ½ C
• Remember, that the circumference of a
circle is 2 π R. So one half of 2 π R would
be π R. Therefore the base of this figure is
π R (the base uses half, and the top of the
figure uses the other half of the
circumference.
• The height of this figure is the radius, or R.
• Therefore, the area of this figure is base
times height, or π R times R, or π R2 which
is also the formula for the area of a circle 
R
B (or π R)
Area of a Circle
• Theorem –As demonstrated with the
rectangle in the previous slide, the area of a
circle is Pi x R2
• Example: How much more pizza is in a 12
inch diameter pizza than a 10 inch pizza?
• First find the radius of each 6 in. and 5 in.
• Area of first pizza = π x (6)2 or 36 π
• Area of second pizza = π x (5)2 or 25 π
• The difference is about 11 π, or 34.55
square inches
Sectors of Circles
• A sector of a circle is a region bounded by an arc
of the circle and the two radii to the arc’s end
points. –In other words, it’s a slice of pie.
• You name the sector by using one arc endpoint,
the center of the circle, and the other arc
endpoint.
• A sector is a fractional part of the area of the
circle.
• Just as we measured the length of an arc by
finding the ratio of the part of the circumference to
the whole of the circumference (arc0/3600), we
find the area of a sector by finding the same ratio
of the part to the whole.
Area of a Sector of a Circle
• Theorem 7.16: The area of a sector of a
circle is the product of the ratio
(Arc0)/360 x the area of the circle (π R2)
• Or
Area of Sector AOB = m(arc AB)/360 x π R2
(Remember, the measure of the arc is the
measure of degrees)
A
B
r
O
Example –Find the area of a Sector
of a Circle
• Find the area of sector
ZOM. Leave answer in
terms of π.
• Area of Sector ZOM =
m(ZOM)/360 x π R2
• = 72/360 x π(20)2
• = 80 π
• The area is 80 π cm2.
Z
720
20 cm
M
O
Check Understanding
• A circle has a diameter of 20 cm. What is
the area of a sector bounded by a 2080
major arc? Round your answer to the
nearest tenth.
• Use the equation m(arc)/360 x π x R2
• = 208/360 x π x (10)2
• = 181.5 cm2
Segment of a Circle
• A part of a circle bounded by an arc and a line
segment joining the arc’s endpoints is called a
Segment of a Circle.
• To find the area of a segment for a minor arc,
draw radii to form a sector.
• The area of the segment equals the area of
the sector minus the area of the triangle
Area of Segment
formed. Area of Sector Area of Triangle
Given a Segment
Make a sector
=
Make a Triangle
Calculate area
Example Find the Area of a
Segment of a Circle
• Find the area of the shaded
segment of the circle. Round to
the nearest tenth.
• Area of sector = m(AB)/360 x
π(R )2
• = 90/360 x Pi(10)2
• = 25 π in2
• Area of Triangle AOB = ½ B x H
• = ½ x 10 x 10
• = 50 in2
• Area of Segment = 25 π – 50
• = 28.5 in2
A
O 10 in.
B
Assignment
• Page 663/64 7-31
• Worksheet 7-7
Unit 10 Quiz 2
Hint: Convert
Percent to a decimal
• Use the following information to calculate the area
of each slice of pie (round to the nearest 10th ):
Car
Owner
Mustang 41%
Corvette 10%
Camaro
25%
G8
8%
BMW 5
5%
Jetta
7%
Viper
4%
Sports Car Owners
1
5%
2
7%
3
4
5
6
4%
41%
8%
25%
Hint: What is the area of the circle?
7
10%
Radius = 15 inches
Circles in the Coordinate Plane
• You can use the Distance Formula to find an equation
of a circle with center point (h,k) and radius r. Chose
(x,y) as any point on a circle, then you can express the
radius r as the distance from point (h,k) to point (x,y).
• Original distance formula:
(x,y)
y
2
2
• d = √ (x2-x1) + (y2-y1)
-same as
r
x2, y2
• Substitute “h” for x1 and “k” for y1
r = √ (x-h)2 + (y-k)2 -Revised Distance
Formula
r2= (x-h)2 + (y-k)2
-Square both sides
(x-h)2 + (y-k)2 = r2 - Re-write in the
“equation of a
circle” format
(h,k)
-same as
x1, y1
0
x
Equation of a Circle
• An equation of a circle with center (h,k)
and radius r is
 (x-h)2 + (y-k)2=r2
• This equation is in standard form. It is
also known as the standard equation of a
circle
• Normally you are given a point (h,k) and a
radius “r”
Example of a Standard Equation of
a Circle
• Write the standard equation of a circle with
center (5, -2) and radius 7
• Use standard form (x-h)2 + (y-k)2=r2
• (x-5)2 + (y-(-2))2 = 72 Substitute variables
• (x-5)2 + (y+2)2 = 49 Simplify
• Write the standard equation of the circle with
center (3,5) and radius 6
• (x-3)2 + (y-5)2 = 36
• Write the standard equation of the circle with
center (-2,-1) and radius √2
• (x+2)2 + (y+1)2 = 2
Using the Center, and another
point on the Circle
• Suppose you have a circle with center (1,-3)
and the circle passes through point (2,2).
Write the standard equation for this circle
• First solve for r
• r = √(x-h)2 + (y-k)2 or √(2-1)2+(2-(-3))2
• = √1 + 25 = √26
• Now use standard form (x-h)2 + (y-k)2=r2
• (x-1)2 + (y-(-3))2 = (√26)2 Substitute
• (x-1)2 + (y+3)2 = 26
Simplify
Another Look
• Using the standard form, determine a
circle’s center and it’s radius.
• (x-7)2 + (y+2)2 = 64
• (x-7)2 + (y-(-2))2 = 82 Put in Standard Form
•
h
k
r
• Therefore the center is (7, -2) and the
radius is 8
Assignment
• Page 800 8-36
• Worksheet 11-5
Unit 10 Quiz 3
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Find the area of a circle with a radius of 12 inches
Find the area of 45% of that circle
Find the area of 67% of that circle
Find the area of 4% of that circle
Find the area bounded by a 88 degree arc of that circle
Find the area bounded by a 129 degree arc of that circle
Find the area of a circle with a diameter of 6 feet
Find the area of 34% of that circle
Find the area bounded by a 230 degree arc of that circle
Find the area bounded by a 310 degree arc of that circle
Tangent Lines In Relation to a
Circle
• What if you drew a circle and
labeled it O
• Then you drew a line which
intersected the circle in only
one point, and labeled it Point A
• Then you drew radius OA
• What seems to be true about
the two angles created by your
line, and radius OA?
• They are Right Angles,
therefore the line is
perpendicular to the radius.
O
A
Tangent to a Circle
• Previously we learned about the tangent ratio in
Right Triangles. Here we will earn about
tangents in relation to circles.
• A tangent to a circle is a line (in the same plane)
that intersects the circle in exactly one point.
• This point (where they intersect, or share one
point) is called the point of tangency.
• You can have tangent lines, tangent rays, or
tangent segments, but they all intersect in
exactly one point (the point of tangency).
Theorem 11.1
• If a line is a tangent to a circle, then the
line is perpendicular to the radius drawn
to the point of tangency.
• Here line AB is perpendicular to segment
P
OP
A
B
O
Example Finding Angle Measures
• Segment MN and Segment
ML are tangent to Circle O.
Find the value of X
• Since the segments are
tangent, Angle L and Angle N
are right angles.
• LMNO is a quadrilateral whose
interior angle measures is
3600
• Therefore, 360 – 90 – 90 –
117 = X0
• X = 630
O
N
1170
X0
M
L
Check Understanding
• ED is tangent to Circle O. Find E
the value of X.
• Because ED is perpendicular to
the radius OD, angle D is a right
angle
• The sum of the interior angles in
a triangle is 180 degrees.
• Therefore, 180 – 90 – 38 = 52
• X = 520
D
X0
.
O
380
Finding the Tangent
• Is ML tangent to Circle N at point L?
Explain
• Determine whether triangle LMN is a
right triangle -because we need to
know if angle L is a right angle
• Does 72 + 242 = 252?
• 625 = 625 YES
• We can conclude that angle L is a
right angle
• Therefore we can conclude that
radius NL is perpendicular to
segment LM
• Therefore the segment is tangent to
Circle N at point L
N
25
M
7
24
L
Circumscribed and Inscribed
Circles
• Previously, we learned that a circle is circumscribed
about a triangle if all the vertices (corners) of the triangle
lie on points of the circle. In this case, the triangle is
inscribed in the circle.
• Similarly, when a circle is inscribed inside a triangle,
then we can say the triangle is circumscribed about the
circle. Each side of the triangle would then be
considered tangent to the circle.
Theorem 11.3
• Look at the inscribed circle below.
• What conclusions could you draw about segment AD
and Segment AF?
• Or segment BD and Segment BE?
• Or segment CF and Segment EC?
• Each pair of segments is congruent
• Theorem 11.3: The two segments tangent to a circle
from one point outside the circle are congruent. Or:
– AD = AF
– BD = BE
– CE = CF
B
D
A
E
F
C
Find the Perimeter
B
• Circle O is inscribed in
triangle ABC
• Find the perimeter of triangle
ABC
• Knowing that inscribed A
circles are tangent, we can
conclude that AF = AD, BE =
BD, and CE = CF
• Therefore, the perimeter of
triangle ABC = 2(10) + 2(15)
+ 2(8) = 66 cm
8 cm
D
E
O
15 cm
F
10 cm
C
Find a Segment Length
B
• Circle O is inscribed in
Triangle ABC
• Triangle ABC has a
perimeter of 88 cm
• Find the length of
segment BE
• Perimeter (88) = 2(17)
+ 2(15) + 2(BE)
• 2(BE) = 88 – 34 – 30
• 2(BE) = 24
• BE = 12 cm
D
E
17 cm
A
F
15 cm
C
Assignment
• Page 767 6-19
• Worksheet 11-1
Unit 10 Quiz 4
1.
Write a standard equation of a circle with a center point of (3,5) and a
radius of 6
2. Write a standard equation of a circle with a center point of (7,2) and a
radius of 9
3. Write a standard equation of a circle with a center point of (3,5) and
goes through point (6,9)
4. Write a standard equation of a circle with a center point of (7,5) and
goes through point (16,9)
5. Write a standard equation of a circle with a center point of (0,0) and
goes through point (6,1)
6. What is the center point and radius of a circle with this standard
equation: x2 + y2 = 169
7. What is the center point and radius of a circle with this standard
equation: (x-3)2 + (y+2)2 = 256
8. Find the area of a circle with a 15 foot diameter (leave answer in pi)
9. Find the area bounded by a 111 degree arc of that circle (in pi)
10. Find the area bounded by a 300 degree arc of that circle (in pi)
Chords
• A segment whose endpoints are on a circle is
called a chord
• A diameter is technically a chord, but normally
when talking about chords, we are talking
about all chords but the diameter
• Segment PQ is a chord P
O
Q
Some Basic Ideas
•
•
•
•
•
Within a circle (or in congruent circles):
Congruent Central Angles have congruent Chords
Congruent Chords have congruent Arcs
Congruent Arcs have Congruent Central Angles
Chords which are equally distant from the center
are congruent
P
A
X0
X0
B
Q
Example
9
• Find the value of a
a
9
9
• Find the value of x
18
18
16
X
36
More Theorems
• In a circle, a diameter that is perpendicular to
a chord bisects the chord, and its arc
• Vice-Versa is true too: If a diameter bisects a
chord, it is perpendicular
• Any perpendicular bisector of a chord goes
through the center of the circle (thus it must
be a diameter)
Example
•
•
•
•
Find the missing length
We are looking for the length of r
r
One length of the right triangle is 3
L
One length is 7 –because the line that goes
through the center of the circle (Segment KN)
is perpendicular to the chord, therefore it
bisects the 14 cm chord to 7 cm)
• Therefore, we use the Pythagorean theorem
• R2 = 32+72
• R = 7.6
K
3
14 N
M
Example
• Find the value of X
• Here, we have the standard right
triangle, so we can use the
Pythagorean Theorem
• 42+ (X/2)2 = 6.82
• (X/2)2 = 6.82-42
• X/2 = √(6.82-42)
• X = 2√(6.82-42)
• X = 2 x 5.5
• X = 11
6.8
4
X
Assignment
• Page 776-777-778 8-10,13-18, 30-32
• Worksheet 11-2
Unit 10 Quiz 5
5 ft
320 Feet
• The distance from home plate to the
wall at a baseball field is 320 feet.
• The warning track is 5 feet wide all
the way around
• The warning track arc is 180 degrees
• The ground crew needs to replace
the warning track gravel
• How much area will they have to
replace (How much area is the
warning track)?
Warning track
(180 degree
arc)
Home Plate
Inscribed Angles
• An inscribed angle is an angle that is on
one side of a circle, and the rays of the
angle extend to the other side of the
circle
• The arc that is between each ray that
forms the angle is called the intercepted
arc
• The measure of an inscribed angle is ½
the measure of the intercepted arc
• The measure of angle B = ½ the
measure of arc ac
• Remember, if it is not given, then the
measure of the arc is found by
measuring the central angle
• If arc AC = 50 degrees, what is the
measure of angle B?
A
B
C
Example
• Find the measure of arc a and
angle b
q
• The measure of the inscribed angle
q is 60 degrees
• The intercepted arc (which is
labeled a) is twice the measure of
the angle, so arc a is 120 degrees
• The measure of the inscribed angle
b is the measure of arc a, plus 300
• Therefore angle b = 120 + 30 =
1500/2=750
m
a
600
n
300
b
p
Theorems
• 2 inscribed angles that intercept the same arc
are congruent
• If an angle is inscribed inside a semicircle (half
of a circle) then it is a right angle (because a
semicircle is 180 degrees, and it is half of that)
• The opposite angles of a quadrilateral that are
inscribed inside a circle are supplementary
Example
• Find the measure of the numbered angle
• Angle 1 is inscribed in a semicircle –it is
90 degrees
• Angle 2 captures the same arc as the 25
degree angle –it is 25 degrees
• Angle 3 and 5 are both inscribed in a
semicircle, they are both 90 degrees
• Angle 6 is ½ the sum of 600 and 800, so it
is 70 degrees
• Angle 4 is ½ the other part of the circle –
there are 360 degrees, minus 60 minus
80= 220, therefore Angle 4 = 220/2 = 110
• Or, you could say 360 degrees in a
quadrilateral, minus 2x90, minus 70 = 110
1
2
250
600
4
3
6
5
800
Theorem
• The measure of an angle
formed by a tangent line and
a chord is ½ the measure of
the intercepted arc
• The measure of angle C = ½
the measure of arc BDC
• Notice: Angle c does not
equal 90 degrees, because
the line is not tangent to the
diameter (or radius).
B
D
C
Unit 10 Quiz 6
1.
2.
3.
4.
5.
6.
7.
8.
How many degrees is angle c?
How many degrees is angle a?
How many degrees is angle b?
How many degrees is angle d?
How many degrees is arc mn?
How many degrees is arc man?
How many degrees is arc pn?
If arc bw is 10 degrees, how many
degrees is arc nb?
9. How many degrees is angle p?
10. How many degrees is arc pd?
300
n
m
300
.
p
b
c
w
a
600
d
Assignment
• Text P. 784-85 6-18,20,21,23-25
• Worksheet 11-3
Angle Measures and Segment Lengths
•
•
•
• Angle 1 is (60+50)/2 = 550
A secant is a line that intersects a circle at two
points
A part of a secant forms a chord inside a circle
The measure of an angle formed by two lines
that intersect INSIDE a circle is ½ the sum of
the measures of the two intercepted arcs
in other words, you take the
60
1
average of 2 angles
What is the measure of angle 1?
0
•
500
Another Theorem
• The measure of an angle formed by two lines
that intersect OUTSIDE the circle is ½ the
difference of the measures of the intersected
arcs
• In other words, instead of adding the 2 angles,
you subtract them and then
1
30
divide by 2 –what is angle 1?
• The measure of angle
100
1 is (1000-300)/2 = 350
0
0
Examples
• Find the value of the variable
950
460
Z0
X
200
900
•(95-Z)/2 = 20
• 95 – Z = 2 x 20 = 40
• -Z = 40 – 95 = -55
• Z = 55
• X = (46+90)/2 = 68
The Ice Cream Cone Problem
• Find the value of X
• Here, all we are given is that the measure of
the angle is 40 degrees
• We know that part of the circle is X degrees
• The other part of the circle would then be
360-X degrees
• Therefore, we can say that (360-X –X)/2 = 40
• 360 – 2X = 80
• -2X = 80 – 360 = -280
• X = -280/-2 = 1400
• What is the measure of the other side of the
cone?
X
400
Theorem
• For a given point and a circle, the product (we
multiply) of the lengths of the two segments
from the point to the circle is constant along
the line through the point and the circle
• axb=cxd
w(w+x) = y(y+z) t2 = y(y+z)
a
d
x
c
w
b
z
t
y
z
y
Example
• Find the value of y
•
•
•
•
6
8(6+8) = 7(7+y)
112 = 49 + 7y
63 = 7y
Y=9
8
7
y
• Find the value of z
• Z2= 8(16+8)
• Z2= 192
• Z = 13.9
z
• Find the value of a
• 6.5 x a = 7 x 3
• 6.5a = 21
• a = 3.2
16
6.5 3
7
a
8
Assignment
• Text P. 794-795 8-13,15-20,24-26
• Worksheet 11-4