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Chapter 10
Properties of Circles
Date: 3/26
Aim: 10.1 Use Properties of Tangents
Homework: Page 655 - 656 #'s (3 - 10, 18,
20, and 22)
Circle
 A set
of all points in a plane that are
equidistant from a given point (the center).
Radius
 Segment
whose endpoints are the center
and any point on the circle.
Chord
 A segment
whose endpoints are on a circle.
Diameter
 A chord
that contains the center of a circle.
Secant
 A line
that intersects a circle in 2 points.
Tangent
 A Line
in a plane of a circle that intersects the
circle in exactly one point, the point of
tangency.
EXAMPLE
1 the line, ray, or segment is best
Tell whether
described as a radius, chord, diameter, secant, or
tangent of C.
a.
a.
AC
AC is a radius because C is the center and A is a
point on the circle.
Tell whether the line, ray, or segment is best
described as a radius, chord, diameter, secant, or
tangent of C.
b.
b.
AB
AB is a diameter because it is a chord that
contains the center C.
Tell whether the line, ray, or segment is best
described as a radius, chord, diameter, secant, or
tangent of C.
c.
c.
DE
DE is a tangent ray because it is contained in a
line that intersects the circle at only one
point.
Tell whether the line, ray, or segment is best
described as a radius, chord, diameter, secant, or
tangent of C.
d.
d.
AE
AE is a secant because it is a line that
intersects the circle in two points.
Use the diagram to find the given lengths.
a.
Radius of
b.
c.
Diameter of A
d.
Diameter of
a.
The radius of
b.
The diameter of A is 6 units.
c.
The radius of
d.
The diameter of B is 4 units.
Radius of
A
B
B
A is 3 units.
B is 2 units.
Theorem
 In
a plane, a line is tangent to a circle if
and only if the line is perpendicular to a
radius of the circle at its endpoint on the
circle.
In the diagram, PT is a radius of
tangent to P ?
P. Is ST
Use the Converse of the Pythagorean Theorem.
Because 122 + 352 = 372,
PST is a right triangle and
ST PT . So, ST is perpendicular to a radius of P
at its endpoint on P. By Theorem 10.1, ST is tangent
to P.
In the diagram, B is a point of tangency. Find the
radius r of C.
You know from Theorem 10.1 that AB BC , so
ABC
is a right triangle. You can use the Pythagorean
Theorem.
AC2 = BC2 + AB2 Pythagorean Theorem
(r + 50)2 = r2 + 802
r2 + 100r + 2500 = r2 + 6400
100r = 3900
r = 39 ft .
Substitute.
Multiply.
Subtract from each side.
Divide each side by 100.
Introduction to tomorrow’s lesson.
Chapter 10
Properties of Circles
Date: 3/27
Aim: 10.1 Use Properties of Tangents
and 10.2 Find Arc Measures
Homework: Page 656 #’s (24 and 26)
Page 661 #’s (3 – 9, 13) ODD ONLY
10.1-10.2 Quiz on Thursday, 3/29
Theorem
 Tangent
segments from a common
external point are congruent.
RS is tangent to C at S and RT is tangent to
Find the value of x.
C at T.
RS = RT
Tangent segments from the same point are
28 = 3x + 4
Substitute.
8=x
Solve for x.
1.
Is DE tangent to
C?
Yes
2.
ST is tangent to Q.Find the value of r.
r=7
3.
Find the value(s) of x.
+3 = x
Vocabulary

Central Angle


Minor Arc


If angle ACB is less than 180°
Major Arc


An angle whose vertex is the
center of the circle.
Points that do not lie on the
minor arc.
Semi Circle

Endpoints are the diameter
Measures
 Measure

The measure of it’s central angle.
 Measure

of a Minor Arc
of a Major Arc
Difference between 360 and the measure of the
minor arc.
Find Arc Measures
Find the measure of each arc of P, where RT is a
diameter.
a.
RS
b.
a.
RS is a minor arc, so mRS = m
b.
RTS is a major arc, so mRTS = 360o – 110o = 250o.
c.
RT is a diameter, so RST is a semicircle, and mRST
= 180o.
RTS
c.
RST
RPS = 110o.
Arc Addition Postulate
 The
measure of an arc formed by two
adjacent arcs is the sum of the measures of
the two arcs.
EXAMPLE 2Find Arc
Measures
A recent survey asked teenagers if they would rather meet
a famous musician, athlete, actor, inventor, or other
person. The results are shown in the circle graph. Find the
indicated arc measures.
a.
mAC
a.
mAC = mAB + mBC
= 29o + 108o
= 137o
b.
b.
mACD
mACD = mAC + mCD
= 137o + 83o
= 220o
Examples
Identify the given arc as a major arc, minor arc, or
semicircle, and find the measure of the arc.
1.
TQ
2.
QRT
minor arc, 120°
major arc, 240°
3. TQR
semicircle, 180°
4.
QS
minor arc, 160°
5.
TS
minor arc, 80°
6.
RST
semicircle, 180°
Congruent Circles

Two circles are congruent if they have the same
radius.

Two arcs are congruent if they have the same
measure and they are arcs of the same circle (or
congruent circles).
Are the red arcs congruent?
Yes
No
Chapter 10
Properties of Circles
Date:
Aim: 10.3 Apply Properties of Chords
Do Now: Quiz Time.
Theorem

In the same circle, two minor arcs are congruent if
and only if their corresponding chords are
congruent.
arcAB  arcDC iff AB  CD
EXAMPLE 1
In the diagram, P
mFG
Q, FG
So, mFG = mJK = 80o.
JK , and mJK = 80o. Find
GUIDED
Try OnPRACTICE
Your Own
Use the diagram of
D.
1. If mAB = 110°, find mBC
mBC = 110°
2. If mAC = 150°, find mAB
mAB = 105°
Theorems
 If
one chord is a perpendicular
bisector of another chord, then
the first chord is a diameter.
 If
one diameter of a circle is
perpendicular to a chord, then
the diameter bisects the chord
and its arc.
EXAMPLE 3
Use the diagram of E to find the length of AC .
Diameter BD is perpendicular to AC .
So, by the Theorem, BD bisects AC ,
and CF = AF.
Therefore, AC = 2 AF = 2(7) = 14.
Try On Your Own
Find the measure of the indicated arc in the diagram.
1.
CD
2.
DE
mCD = 72°
mCD = mDE.
mDE = 72°
3.
CE
mCE = mDE + mCD
mCE = 72° + 72° = 144°
Theorem
 In
the same circle, two chords are congruent if
and only if they are equidistant from the center.
In the diagram of C, QR = ST = 16. Find CU.
CU = CV
2x = 5x – 9
x=3
So, CU = 2x = 2(3) = 6.
Use Theorem.
Substitute.
Solve for x.
Try On Your Own
In the diagram, suppose ST = 32, and CU = CV = 12. Find the
given length.
1.
QR
QR = 32
2. UR
UR = 16
3. The radius of
C
The radius of
C = 20
Homework
 Page

667 – 668
#’s (3 – 11 and 18 - 20)
 Study
3/28
for 10.3 – 10.4 Quiz on Monday
Chapter 10
Properties of Circles
Date:
Aim: 10.4 Use Inscribed Angles and Polygons
Do Now:
Measure of an Inscribed Angle Theorem

The measure of an inscribed angle is one half the
measure of its intercepted arc.
A
inscribed
angle D
intercepted
arc
●C
B
1 
mADB  mAB
2
An inscribed angle is an angle whose vertex is on a
circle and whose sides contain chords of the circle.
Example
S
T
R
Q
mQTS = 2m QRS = 2 (90°) = 180°
Theorem

If two inscribed angles of a circle intercept the same
arc, then the angles are congruent.
E
D
F
C
mEDF  mECF
Try On Your Own
Find the measure of the red arc or angle.
1.
2.
3.
m
G = 1 mHF = 1 (90o) = 45o
2
2
mTV = 2m
ZYN
ZXN
U = 2 38o = 76o.
ZXN
72°
Inscribed Polygons
 A polygon
is inscribed if all of its vertices lie on
a circle.

Circle containing the vertices is a Circumscribed
Circle.
Theorem
 A right
triangle is inscribed in a circle if and
only if the hypotenuse is a diameter of the
circle.
●C
Theorem
 A quadrilateral
is inscribed in a circle if and
only if its opposite angles are supplementary.
●C
y = 105°
x = 100°
Try On Your Own
Find the value of each variable.
1.
2.
y = 112
x = 98
c = 62
x = 10
Homework
 Page

676
#’s (3 – 8, 13 – 15, 18)
 Study
for 10.3 – 10.4 Quiz on Monday,
3/28
 Quiz on 10.5 – 10.7 on Friday, 4/1
 Chapter 10 Test Tuesday April 5th
Chapter 10
Properties of Circles
Date:
Aim: 10.5: Apply Other Angle Relationships in
Circles
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.
1 
mBAE  mAB
2
1  
mBAD  mBCA
2
Line m is tangent to the circle. Find the measure of the red
angle or arc.
a. m
1
1 = 2 (130o) = 65o
o
b. m KJL = 2 (125 ) = 250o
Try On Your Own
Find the indicated measure.
m
1
1 = 2 (210o) = 105o
o
m RST = 2 (98 )
o
m XY = 2 (80 )
= 196o
= 160o
Angles Inside the Circle Theorem

If two chords intersect inside a circle, then the
measure of each angle is one half the sum of the
measures of the arcs intercepted by the angle and its
vertical angles.
x°
xo =
1 (mBC + mDA)
2
Find the value of x.
The chords JL and KM intersect inside
the circle.
xo =
1 (mJM + mLK)
2
Use Theorem 10.12.
xo =
1
(130o + 156o)
2
Substitute.
xo = 143
Simplify.
Angles Outside the Circle Theorem

If a tangent and a secant, two tangents, or two
secants intersect outside a circle, then the measure of
the angle formed is one half the difference of the
measures of the intercepted arcs.
D

1  
mP  ADB  AB
2

Find the value of x.
The tangent CD and the secant CB
intersect outside the circle.
m
BCD =
xo =
1 (mAD – mBD)
2
Use Theorem 10.13.
1
(178o – 76o)
2
Substitute.
x = 51
Simplify.
Try On Your Own
Find the value of the variable.
5.
y = 61o
6.
o
a = 104
xo
253.7o
Homework
 Page

683 - 684
#’s (7 – 12)
 Chapter
10 Test Wednesday April 6th
Chapter 10
Properties of Circles
Date:
Aim: 10.6: Find Segment Lengths in Circles
Do Now:
Segments of the Chords
 If
two chords intersect in the interior of the
circle, then the product of the lengths of the
segments of one chord is equal to the product
of the lengths of the segments of the other
chord.
Find ML and JK.
EXAMPLE 1
NK NJ = NL NM
x (x + 4) = (x + 1)
(x + 2)
x2 + 4x = x2 + 3x + 2
4x = 3x + 2
x = 2
ML = ( x + 2 ) + ( x + 1)
Use Theorem
Substitute.
Simplify.
Subtract x2 from each side.
Solve for x.
JK = x + ( x + 4)
= 2+2+2+1
= 2+2+4
= 7
= 8
Segments of Secants Theorem
 If
two secant segments share the same
endpoint outside a circle, then the product of
the lengths of one secant segment and its
external segment equals the product of the
lengths of the other secant segment and its
external segment.
Find x.
RQ
RP = RS
4 (5 + 4) = 3
RT
(x + 3)
36 = 3x + 9
9 = x
Use Theorem 10.15.
Substitute.
Simplify.
Solve for x
The correct answer is D.
Try On Your Own.
Find the value(s) of x.
13 = x
x = 8
3 = x
Segments of Secants and Tangents
Theorem
 If
a secant segment and a tangent segment
share an endpoint outside a circle, then the
product of the lengths of the secant segment
and its external segment equals the square of
the length of the tangent segment.
Use the figure at the right to find RS.
RQ2 = RS RT
162 = x
Use Theorem.
(x + 8)
Substitute.
= x2 + 8x
Simplify.
0
= x2 + 8x – 256
Write in standard form.
x
= –8 +
x
= – 4 + 4 17
256
82 – 4(1) (– 256)
2(1)
So, x = – 4 + 4 17
Use quadratic formula.
Simplify.
12.49, and RS
12.49
Try On Your Own.
Find the value of x.
1.
2.
x
3.
=2
x
= 24
5
x
=8
Try On Your Own.
Then find the value of x.
1.
= – 7 + 274
x
2.
x
= 8
x
= 16
3.
Homework
 Page

692 - 693
#’s ()
 Chapter
10 Test Wednesday April 6th
Chapter 10 Properties of Circles
Date:
Aim: 10.7: Write and Graph Equations of Circles
Do Now: Take out Homework from Tuesday
night.
Find the value of x for the problems below.
2.
1.
x
= 2
x
= 24
5
Standard Equation of a Circle
x  h    y  k 
2
2
r
2
Why?
a b c
2
●
(x, k)
2
2
( x  h)  ( y  k )  r
2
2
2
( x  h)  ( y  k )  r
2
2
Write the equation of the circle shown.
The radius is 3 and the center is at the origin.
x2 + y2 = r2
Equation of circle
x2 + y2 = 32
Substitute.
x2 + y2 = 9
Simplify.
The equation of the circle is x2 + y2 = 9
Write the standard equation of a circle with center (0, –9) and
radius 4.2.
4.2
C●
(0, -9)
(x – h)2 + ( y – k)2 = r2
Standard equation of a circle
(x – 0)2 + ( y – (–9))2 = 4.22
Substitute.
x2 + ( y + 9)2 = 17.64
Simplify.
Try On Your Own.
Write the standard equation of the circle with the given
center and radius.
1. Center (0, 0), radius 2.5
x2 + y2 = 6.25
2. Center (–2, 5), radius 7
(x + 2)2 + ( y – 5)2 = 49
The point (–5, 6) is on a circle with center (–1, 3). Write the
EXAMPLE
3
standard
equation
of the circle.
Steps:
1. Find values of h, k, and r by using the distance
formula.
2
2
d
x2  x1    y2  y1 
2. Substitute your values into the equation for a
circle.
r =
=
[–5 – (–1)]2 + (6 – 3)2
(–4)2 + 32
= 5
Substitute (h, k) = (–1, 3) and r = 5 into the equation of the circle.
(x – h)2 + (y – k)2 = r2
[x – (–1)]2 + (y – 3)2
(x +1)2 + (y – 3)2
= 52
=
25
(x +1)2 + (y – 3)2 = 25.
Try On Your Own.
1. The point (3, 4) is on a circle whose center is (1, 4). Write the
standard equation of the circle.
The standard equation of the circle is
(x – 1)2 + (y – 4)2 = 4.
2. The point (–1 , 2) is on a circle whose center is (2, 6).
Write the standard equation of the circle.
The standard equation of the circle is
(x – 2)2 + (y – 6)2 = 25.
Graph A Circle
The equation of a circle is (x – 4)2 + (y + 2)2 = 36.
GRAPH
Rewrite the equation to find the center
and radius.
(x – 4)2 + (y +2)2
= 36
(x – 4)2 + [y – (–2)]2 = 62
The center is (4, –2) and the radius is 6.
Try On Your Own.
1. The equation of a circle is (x – 4)2 + (y + 3)2 = 16.
Graph the circle.
6. The equation of a circle is (x + 8)2 + (y + 5)2 = 121.
Graph the circle.
Homework
 BRING
WORKBOOK TO CLASS tomorrow
 Chapter 10 Test Wednesday April 6th