P
DIAMETER:Distance across the circle through its centerAlso known as the longest chord.
P
RADIUS:
Distance from the center to point on circle
Formula
Radius = ½ diameteror
Diameter = 2r
D = ?
r = ?
r = ? D = ?
Secant Line:intersects the circle at exactly TWO points
a LINE that intersects the circle exactly ONE time
Tangent Line:
Forms a 90°angle with one radius
Point of Tangency: The point where the tangent intersects the circle
Name the term that best describes the notation.
Secant
Radius
DiameterChord
Tangent
Central Angles
An angle whose vertex is at the center of the circle
P
E
F
D
Semicircle: An Arc that equals 180°
EDF
To name: use 3 letters
THINGS TO KNOW AND REMEMBER ALWAYS
A circle has 360 degrees
A semicircle has 180 degrees
Vertical Angles are CONGRUENT
Linear Pairs are SUPPLEMENTARY
Formulameasure Arc = measure
Central Angle
m AB
m ACB
m AE
A
B
C
Q96
E=
=
=
96°
264°
84°
Find the measures. EB is a diameter.
Tell me the measure of the following arcs.
AC is a diameter.
80
10040
140A
B
C
D
Rm DAB =
m BCA =
240
260
Using Properties of Tangents
HK and HG are tangent to F. Find HG.
HK = HG
5a – 32 = 4 + 2a
3a – 32 = 4
2 segments tangent to from same ext. point segments .
Substitute 5a – 32 for HK and 4 + 2a for HG.
Subtract 2a from both sides.
3a = 36
a = 12
HG = 4 + 2(12)
= 28
Add 32 to both sides.
Divide both sides by 3.
Substitute 12 for a.
Simplify.
Applying Congruent Angles, Arcs, and Chords
TV WS. Find mWS.
9n – 11 = 7n + 11
2n = 22
n = 11
= 88°
chords have arcs.
Def. of arcs
Substitute the given measures.
Subtract 7n and add 11 to both sides.
Divide both sides by 2.
Substitute 11 for n.
Simplify.
mTV = mWS
mWS = 7(11) + 11
TV WS
Example 3B: Applying Congruent Angles, Arcs, and Chords
C J, and mGCD mNJM. Find NM.
GD = NM
arcs have chords.GD NM
GD NM GCD NJM
Def. of chords
Find QR to the nearest tenth.
Step 2 Use the Pythagorean Theorem.
Step 3 Find QR.
PQ = 20 Radii of a are .
TQ2 + PT2 = PQ2
TQ2 + 102 = 202
TQ2 = 300TQ 17.3
QR = 2(17.3) = 34.6
Substitute 10 for PT and 20 for PQ.Subtract 102 from both sides.Take the square root of both sides.
PS QR , so PS bisects QR.
Step 1 Draw radius PQ.
The circle graph shows the types of cuisine available in a city. Find mTRQ.
158.4
Inscribed Angle
Inscribed Angle = intercepted Arc/2
160
80
The inscribed angle is half of the intercepted angle
120
x
y
Find the value of x and y.
= 120
= 60
In J, m3 = 5x and m 4 = 2x + 9.Find the value of x.
3
Q
D
JT
U
4
5x = 2x + 9
x = 3
3x = + 9
4x – 14 = 90
H
K
GN
Example 4
In K, GH is a diameter and mGNH = 4x – 14. Find the value of x.
x = 26
4x = 104
z
2x + 18
85
2x +18 + 22x – 6 = 180
x = 7
z + 85 = 180z = 95
Example 5 Solve for x and z.
22x – 6
24x +12 = 18024x = 168
1. Solve for arc ABC
2. Solve for x and y.
244
x = 105y = 100
Vertex is INSIDE the Circle NOT at the Center
Arc+ArcANGLE =
2
Ex. 1 Solve for x
X
8884
x = 100
180 – 88
92
8492
2x
184 84 x
Ex. 2 Solve for x.
45
93
xº
89x = 89
360 – 89 – 93 – 45
133
133 452
x
Vertex is OUTside the Circle
Large Arc Small ArcANGLE =
2
x
Ex. 3 Solve for x.
65°
15°
x = 25
65 152
x
x
Ex. 4 Solve for x.
27°
70°
x = 16
7027
2x
54 70 x
x
Ex. 5 Solve for x.
260°
x = 80
360 – 260
100
260 1002
x
Warm up: Solve for x
18◦
1.)
x
124◦
70◦
x
2.)
3.)
x
260◦
20◦110◦ x
4.)
53 145
8070
Circumference, Arc Length, Area, and Area of Sectors
Find the EXACT circumference.
28 ftC 1. r = 14 feet
2. d = 15 miles
15 milesC
2 14C
15C
Ex 3 and 4: Find the circumference. Round to the nearest tenths.
89.8 mmC 103.7 ydC
2 14.3C 33C
Arc LengthThe distance along the curved line
making the arc (NOT a degree amount)
Arc Length
measure of arc
Arc Length 2360
r
Ex 5. Find the Arc LengthRound to the nearest hundredths
8m
70
Arc Length 9.7= 7 m
measure of arc
Arc Length 2360
r
70Arc Length 2 8
360
Ex 6. Find the exact Arc Length.
Arc Length 10
in3
=
measure of arc
Arc Length 2360
r
120Arc Length 2 5
360
Ex 7. What happens to the arc length if the radius were to be doubled? Halved?
20Doubled
35
Halved 3
measure of arc
Arc Length 2360
r
Area of CirclesThe amount of space occupied.
r A = pr2
Find the EXACT area.
2841 ftA 8. r = 29 feet
9. d = 44 miles
2484 miA
229A
2442
A
10 and 11Find the area. Round to the nearest tenths.
2181.5 ydA 22206.2 cmA
27.6A
2532
A
Area of a Sectorthe region bounded by two radii of the
circle and their intercepted arc.
Area of a Sector
2measure of arc
360A r
Example 12Find the area of the sector to the nearest hundredths.
A 18.85 cm2
606 cm
Q
R
2606
360A
Example 13 Find the exact area of the sector.
6 cm
120
7 cm
Q
R
249A cm
3
21207
360A
Area of minor segment =
(Area of sector) – (Area of triangle)
12 yd
2 1
Area of minor segment =360 2
mRQr b hR
Q290 1
= (12) (12)(12)360 2
=113.10 722Area of minor segment =41.10yd
Example 14