Circles in Analytic Geometry
SHS Analytic Geometry
Unit 5
Objectives/Assignment
Week 1: G.GPE.1; G.GPE.4
Students should be able to derive the formula for a circle
given the Pythagorean Theorem
Students should be able to derive the equation of a circle
given its center and radius.
Students should be able to complete the square to find the
center and radius of a circle given by an equation.
Students should prove properties involving circles on the
coordinate plane.
Students should be able to solve circle Application
Problems.
What is a circle?
A circle is the set of all points in a plane equidistant from
a fixed point.
“
Equi
” means same,
so equidistant
means the same
distance.
The
fixed point
is
called the center.
equi
fixed
point
EQ: What does a triangle have to
do with a circle?
http://learni.st/learnings/28476-
the-equation-of-a-basic-
circle?board_id=3075
CIRCLE TERMS
(x – h)² + (y – k)² = r²
(h, k )
r
r
C
=
(
h
,
k
)
Definition: A circle is an infinite
number of points a set distance
away from a center
Finding Equations of Circles
You can write an
equation of a circle in a
coordinate plane if you
know its radius and the
coordinates of its center.
Finding Equations of Circles
Suppose the radius is r
and the center is (h, k).
Let (x, y) be any point on
the circle. The distance
between (x, y) and (h, k)
is r, so you can use the
Distance Formula. (Told
you it wasn’t going
away).
Finding Equations of Circles
Square both sides to
find the standard
equation of a circle
with radius r and
center (h, k).
(x – h)
2
+ (y – k)
2
= r
2
If the center is at the
origin, then the
standard equation is
x
2
+ y
2
= r
2
.
Equation of a circle
Use the distance formula
to determine the
equation of a circle
EX 1 Write an equation of a circle with
center (3, -2) and a radius of 4.
EX 2 Write an equation of a circle with
center (-4, 0) and a
diameter
of 10.
EX 3 Write an equation of a circle with
center (2, -9) and a
radius of
.
EX 4 Find the coordinates of the center and
the measure of the radius.
The center is
The radius is
The equation is
(0, 0)
12
x
2
+ y
2
= 144
The center is
The radius is
The equation is
(1, -3)
7
(x – 1)
2
+ (y + 3)
2
= 49
(x – 3)
2
+ (y – 2)
2
= 9
Center
(3, 2)
Radius of
3
Parameters of circle
Center: (h,k)
The fixed point described in
the definition of a circle
Radius: r
The distance from the center
of the circle to any point on
the circle
r
(h,k)
Ex. 1: Writing a Standard Equation of a
Circle
Write the standard equation of the circle with a
center at (-4, 0) and radius 7.1
(x – h)
2
+ (y – k)
2
= r
2
Standard equation of a circle.
[(x – (-4)]
2
+ (y – 0)
2
= 7.1
2
Substitute values.
(x + 4)
2
+ (y – 0)
2
= 50.41
Simplify.
Ex. 2: Writing a Standard Equation of a Circle
The point (1, 2) is on a circle whose center is (5, -1).
Write a standard equation of the circle.
(x – h)
2
+ (y – k)
2
= r
2
Standard equation of a circle.
[(x – 5)]
2
+ [y –(-1)]
2
= 5
2
Substitute values.
(x - 5)
2
+ (y + 1)
2
= 25
Simplify.
Graphing Circles
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Ex. 3: Graphing a circle
T
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c
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r
c
l
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i
s
(
x
+
2
)
2
+
(
y
-
3
)
2
=
9
.
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c
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.
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s
.
(
x
+
2
)
2
+
(
y
-
3
)
2
=
9
[
x
–
(
-
2
)
]
2
+
(
y
–
3
)
2
=
3
2
T
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n
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r
i
s
(
-
2
,
3
)
a
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s
i
s
3
.
Ex. 3: Graphing a circle
To graph the circle, place
the point of a compass at
(-2, 3), set the radius at 3
units, and swing the
compass to draw a full
circle.
Ex. 2: Writing a Standard Equation of a Circle
The point (1, 2) is on a circle whose center is (5, -1).
Write a standard equation of the circle.
r = 5
Use the Distance Formula
Substitute values.
Simplify.
Simplify.
Addition Property
Square root the result.
Given the graph of a circle,
state its equation
Use the center and the
point (4,4) to find the
radius.
To write the equation of a
To write the equation of a
circle you must know the
circle you must know the
center and the radius.
center and the radius.
From our graph we
From our graph we
see that the center is
see that the center is
at (2,1).
at (2,1).
Given the graph of a circle,
state its equation
Center (2,1)
WRITE and GRAPH
A) write the equation of the circle in
standard form
x
² + y² - 4x + 8y + 11 = 0
Group the x and y terms
x
² - 4x + y² + 8y + 11 = 0
Complete the square for x/y
x
² - 4x +
4
+ y² + 8y +
16
= -11 +
4
+
16
(x – 2)
² + (y + 4)² = 9
YAY! Standard Form!
•
B) GRAPH
•
Plot Center (2,-4)
•
Radius = 3
WRITE and GRAPH
A) write the equation of the circle in
standard form
4x
² + 4y² + 36y + 5 = 0
Group the x and y terms
4x
² + 4y² + 36y + 5 = 0
Complete the square for x/y
4x
² + 4(y² + 9y) = -5
4x
² + 4(y² + 9y +
81/4
) = -5 +
81
4x
² + 4(y + 9/2)² = 76
x
² + (y + 9/2)² = 19
YAY! Standard Form!
•
B) GRAPH
•
Plot Center (0 , -9/2)
•
Radius =
√19 = 4.5
WRITING EQUATIONS
Write the EQ of a circle that has a center of (-5,7) and passes
through (7,3)
Plot your info
Need to find values for h, k,
and r
(h , k) = (-5 , 7)
How do we find r?
Use distance formula with C
and P.
Plug into formula
(x – h)² + (y – k)² = r²
(x + 5)² + (y – 7)² = (4
√10)²
(x + 5)² + (y – 7)² = 160
C =
(-5,7)
P =
(7,3)
radius
Let’s Try One
Write the EQ of a circle that has endpoints of the diameter at (-
4,2) and passes through (4,-6)
A =
(-4,2)
B =
(4,-6)
radius
•
Plot your info
•
Need to find values for h, k, and
r
•
How do we find (h,k)?
•
Use midpoint formula
•
(h , k) = (0 , -2)
•
How do we find r?
•
Use dist form with C and B.
•
Plug into formula
•
(x – h)² + (y – k)² = r²
•
(x)² + (y + 2)² = 32
Hint: Where is the center? How
do you find it?
Suppose the equation of a circle is (x – 5)
² +
(y + 2)² = 9
Write the equation of the new circle given that:
A) The center of the circle moved up 4 spots and left 5:
•
(x – 0)
² +
(y – 2)² = 9
Center moved from (5,-2)
(0,2)
B) The center of the circle moved down 3 spots and right 6:
•
(x – 11)
² +
(y + 5)² = 9
Center moved from (5,-2)
(11,-5)
Find the center and radius of the circle with equation
(
x
+ 4)
2
+ (
y
– 2)
2
= 36.
The center of the circle is (–4, 2). The radius is 6.
(
x
–
h
)
2
+ (
y
–
k
)
2
=
r
2
Use the standard form.
(
x
+
4
)
2
+ (
y
–
2
)
2
= 36
Write the equation.
(
x
– (–4))
2
+ (
y
– 2)
2
= 6
2
Rewrite the equation in
standard form.
h
=
–4
k
=
2
r
= 6
Find
h
,
k
, and
r
.
Graph (
x
– 3)
2
+ (
y
+ 1)
2
= 4.
(
x
–
h
)
2
+ (
y
–
k
)
2
=
r
2
Find the center and radius
of the circle.
(
x
–
3
)
2
+ (
y
– (
–1
))
2
= 4
h
=
3
k
=
–1
r
2
= 4, or
r
= 2
Equation of a Circle
Writing the Equation of a Circle
1.
Group x terms together, y-terms together, and
move constants to the other side
2.
Complete the square for the x-terms
Remember that whatever you do to one side, you must
also do to the other
3.
Complete the square for the y-terms
Remember that whatever you do to one side, you must
also do to the other
Example: Write the equation and find
the center and radius length of :
Group terms
Complete the square
You try!!:
Write the equation and find
the center and radius length of :
THINK ABOUT IT
Find the center, the length of the radius, and write
the equation of the circle if the endpoints of a
diameter are (-8,2) and (2,0).
Center
: Use midpoint
formula!
Length
: use distance formula
with radius and an endpoint
Equation
:
Put it all together
Ex. 4: Applying Graphs of Circles
A bank of lights is arranged over a stage. Each
light illuminates a circular area on the stage. A
coordinate plane is used to arrange the lights,
using the corner of the stage as the origin. The
equation (x – 13)
2
+ (y - 4)
2
= 16 represents one of
the disks of light.
A. Graph the disk of light.
B. Three actors are located as follows: Henry is at
(11, 4), Jolene is at (8, 5), and Martin is at (15, 5).
Which actors are in the disk of light?
Ex. 4: Applying Graphs of Circles
1.
Rewrite the equation to find the center and
radius.
(x – h)
2
+ (y – k)
2
= r
2
(x - 13)
2
+ (y - 4)
2
= 16
(x – 13)
2
+ (y – 4)
2
= 4
2
The center is at (13, 4) and the radius is 4. The circle
is shown on the next slide.
Ex. 4: Applying Graphs of Circles
1.
Graph the disk of light
The graph shows that Henry and Martin are both in the
disk of light.
Ex. 4: Applying Graphs of Circles
A bank of lights is arranged over a stage. Each
light
r = 5
Use the Distance Formula
Substitute values.
Simplify.
Simplify.
Addition Property
Square root the result.
Example 3: Radio Application
An amateur radio operator wants to build a
radio antenna near his home without using
his house as a bracing point. He uses three
poles to brace the antenna. The poles are to
be inserted in the ground at three points
equidistant from the antenna located at
J
(4,
4),
K
(–3, –1), and
L
(2, –8). What are the
coordinates of the base of the antenna?
Step 1
Plot the three given
points.
Step 2
Connect
J
,
K
, and
L
to
form a triangle.
Example 3 Continued
Step 3
Find a point that is equidistant from the
three points by constructing the perpendicular
bisectors of two of the sides of
∆
JKL.
The perpendicular bisectors of the sides of
∆
JKL
intersect at a point that is equidistant from
J
,
K
,
and
L
.
The intersection of the perpendicular
bisectors is
P
(3, –2).
P
is the center
of the circle that passes through
J
,
K
,
and
L
.
The base of the antenna is at
P
(3, –2).
Check It Out!
Example 3
What if…?
Suppose the coordinates of the three
cities in Example 3 (p. 801) are
D
(6, 2) ,
E
(5, –5),
and
F
(-2, -4). What would be the location of the
weather station?
Step 1
Plot the three given
points.
Step 2
Connect
D
,
E
, and
F
to
form a triangle.
Check It Out!
Example 3 Continued
Step 3
Find a point that is equidistant from the
three points by constructing the perpendicular
bisectors of two of the sides of
∆
DEF.
The perpendicular bisectors of the sides of
∆
DEF
intersect at a point that is equidistant from
D
,
E
,
and
F
.
The intersection of the perpendicular
bisectors is
P
(2, –1).
P
is the center
of the circle that passes through
D
,
E
, and
F
.
The base of the antenna is
at
P
(2, –1).
Lesson Quiz: Part III
5.
A carpenter is planning to build a circular gazebo
that requires the center of the structure to be
equidistant from three support columns located
at
E
(
–2, –4
)
,
F
(
–2, 6
)
, and
G
(
10, 2
)
.
What are the coordinates for the location of the
center of the gazebo?
(
3, 1
)
Example 6
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P
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K
E
Three receiving stations are located on a
coordinate plane at points (1, 4), (–
3, –
1), and
(5, 2). The distance from the earthquake
epicenter to each station should be 2 units, 5
units, and 4 units respectively.
S
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(
1
,
2
)
.
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x
=
1
a
n
d
y
=
2
.
Example 6
L
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P
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A
:
Example 6
L
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P
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t
a
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B
:
Example 6
L
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C
:
Example 6
L
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P
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O
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Q
U
A
K
E
Three receiving stations are located on a
coordinate plane at points (1, 4), (–
3, –
1), and
(5, 2). The distance from the earthquake
epicenter to each station should be 2 units, 5
units, and 4 units respectively.
The point (1, 2) does lie on all three graphs;
thus, we can conclude that the epicenter is
at (1, 2).
Explore the fundamentals of circles in analytic geometry by learning how to derive circle formulas, equation determination, completing the square, proving circle properties, and solving application problems. Discover the definition of a circle, its relation to triangles, and the terms, equations, and forms associated with circles. Learn how to find equations of circles using radius and center coordinates, and understand the standard equation of a circle. Dive into practical examples and applications to solidify your understanding of circles in analytical geometry.
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Presentation Transcript
SHS Analytic Geometry Unit 5
Objectives/Assignment Week 1: G.GPE.1; G.GPE.4 Students should be able to derive the formula for a circle given the Pythagorean Theorem Students should be able to derive the equation of a circle given its center and radius. Students should be able to complete the square to find the center and radius of a circle given by an equation. Students should prove properties involving circles on the coordinate plane. Students should be able to solve circle Application Problems.
What is a circle? A circle is the set of all points in a plane equidistant from a fixed point. equi fixed point Equi means same, so equidistant means the same distance. The fixed point is called the center.
EQ: What does a triangle have to do with a circle? http://learni.st/learnings/28476- the-equation-of-a-basic- circle?board_id=3075
Definition: A circle is an infinite number of points a set distance away from a center CIRCLE TERMS EQUATION FORM CENTER (x h) + (y k) = r r (h, k ) C=(h , k) RADIUS r + 2 + 2 x x y y MIDPOINT FORMULA DISTANCE FORMULA , 1 2 1 2 y ( + 2 2 x ( x ) y ) 2 1 2 1
Finding Equations of Circles 6 y You can write an equation of a circle in a coordinate plane if you know its radius and the coordinates of its center. (x, y) 4 (h, k) 2 x 5 10 15 -2
Finding Equations of Circles Suppose the radius is r and the center is (h, k). Let (x, y) be any point on the circle. The distance between (x, y) and (h, k) is r, so you can use the Distance Formula. (Told you it wasn t going away). 6 y (x, y) 4 (h, k) 2 x 5 10 15 -2
Finding Equations of Circles k y h x = + ) ( ) ( 2 2 r 6 y Square both sides to find the standard equation of a circle with radius r and center (h, k). (x h)2 + (y k)2 = r2 If the center is at the origin, then the standard equation is x2 + y2 = r2. (x, y) 4 (h, k) 2 x 5 10 15 -2
Equation of a circle Use the distance formula to determine the equation of a circle (x, y) (h, k) ( ) and ( (x, ) 2 2 = + d x x y y 2 1 2 1 Using (h, k) y), we get ( ) ( ) 2 2 = + d x h y k The distance is d actually equal to the radius of the circle, so we get ( ) ( ) ( ) ( ) . 2 2 2 2 = + = + 2 or r x h y k r x h y k
EX 1 Write an equation of a circle with center (3, -2) and a radius of 4. h k r ( ) ( ) 2 2 2 x h + y k r = ( ) 2 ( ) ( ) 2 2 x 3 + y 2 4 = ( ) ( ) 2 2 x 3 + y 2 16 + =
EX 2 Write an equation of a circle with center (-4, 0) and a diameter of 10. h k ( ) + x h ( ) ( ) 4 x ( ) 4 x + 2r 2 = ( + +y ) 2 2 y ( k r 2 ) 2 2 y 0 5 = 25 2 2 =
EX 3 Write an equation of a circle with center (2, -9) and a radius of . h k ( ) + x h 11 ) r 2 ( 2 2 y k r = ( ) 2 2 ( ) ) ( ) 2 x 2 y 9 11 + = ( ( ) 2 2 x 2 + y+9 11 =
EX 4 Find the coordinates of the center and the measure of the radius. Opposite signs! ( ) ( 6 + x y + ) 2 2 2 3 25 = Take the square root! ( , ) 6 3 Radius 5
5. Find the center, radius, & equation of the circle. (0, 0) The center is 12 The radius is The equation is x2 + y2 = 144
6. Find the center, radius, & equation of the circle. (1, -3) The center is The radius is 7 The equation is (x 1)2 + (y + 3)2 = 49
7. Graph the circle, identify the center & radius. (x 3)2 + (y 2)2 = 9 Center (3, 2) Radius of 3
Parameters of circle Center: (h,k) The fixed point described in the definition of a circle (h,k) Radius: r The distance from the center of the circle to any point on the circle r
Ex. 1: Writing a Standard Equation of a Circle Write the standard equation of the circle with a center at (-4, 0) and radius 7.1 (x h)2 + (y k)2 = r2 Standard equation of a circle. Substitute values. [(x (-4)]2 + (y 0)2 = 7.12 Simplify. (x + 4)2 + (y 0)2 = 50.41
Ex. 2: Writing a Standard Equation of a Circle The point (1, 2) is on a circle whose center is (5, -1). Write a standard equation of the circle. (x h)2 + (y k)2 = r2 Standard equation of a circle. Substitute values. [(x 5)]2 + [y (-1)]2 = 52 Simplify. (x - 5)2 + (y + 1)2 = 25
Graphing Circles If you know the equation of a circle, you can graph the circle by identifying its center and radius.
Ex. 3: Graphing a circle (x+2)2 + (y-3)2 = 9 [x (-2)]2 + (y 3)2=32 The center is (-2, 3) and the radius is 3. The equation of a circle is (x+2)2 + (y-3)2 = 9. Graph the circle. First rewrite the equation to find the center and its radius.
Ex. 3: Graphing a circle To graph the circle, place the point of a compass at (-2, 3), set the radius at 3 units, and swing the compass to draw a full circle. 6 4 2 -5 5 -2
Ex. 2: Writing a Standard Equation of a Circle The point (1, 2) is on a circle whose center is (5, -1). Write a standard equation of the circle. + 2 2 r = Use the Distance Formula ( ) ( ) x x y y 2 1 2 1 Substitute values. ) 1 + ) 2 2 2 5 ( ( 1 r = r = Simplify. + ) 3 2 2 ) 4 ( 16+ 25 ( 9 r = Simplify. r = Addition Property Square root the result. r = 5
Given the graph of a circle, state its equation To write the equation of a circle you must know the center and the radius. From our graph we see that the center is at (2,1). ( ) ( ) 2 2 = + 4 2 4 1 r = + 2 2 2 3 r = + 4 9 r = 13 r Use the center and the point (4,4) to find the radius. (4, 4) (2, 1)
Given the graph of a circle, state its equation = 13 r Center (2,1) The equation = x of the y circle is (4, 4) ( ) ( ) ( ) ( ) 2 2 + 2 r h k ( 13 ( 2 ( 1 ) 2 2 2 = + 13 = 2 1 x y (2, 1) ) )2 2 + x y
WRITE and GRAPH A) write the equation of the circle in standard form x + y - 4x + 8y + 11 = 0 Group the x and y terms x - 4x + y + 8y + 11 = 0 Complete the square for x/y x - 4x + 4 + y + 8y + 16 = -11 + 4 + 16 (x 2) + (y + 4) = 9 YAY! Standard Form! B) GRAPH Plot Center (2,-4) Radius = 3
WRITE and GRAPH A) write the equation of the circle in standard form 4x + 4y + 36y + 5 = 0 Group the x and y terms 4x + 4y + 36y + 5 = 0 Complete the square for x/y 4x + 4(y + 9y) = -5 4x + 4(y + 9y + 81/4) = -5 + 81 4x + 4(y + 9/2) = 76 x + (y + 9/2) = 19 YAY! Standard Form! B) GRAPH Plot Center (0 , -9/2) Radius = 19 = 4.5
WRITING EQUATIONS Write the EQ of a circle that has a center of (-5,7) and passes through (7,3) Plot your info Need to find values for h, k, and r (h , k) = (-5 , 7) How do we find r? Use distance formula with C and P. C = (-5,7) P = (7,3) Plug into formula (x h) + (y k) = r (x + 5) + (y 7) = (4 10) (x + 5) + (y 7) = 160
Lets Try One Write the EQ of a circle that has endpoints of the diameter at (- 4,2) and passes through (4,-6) Plot your info Need to find values for h, k, and r How do we find (h,k)? Use midpoint formula 2 2 A = (-4,2) + + 4 4 2 6 = C , (h , k) = (0 , -2) How do we find r? Use dist form with C and B. 32 Dist = = B = (4,-6) 4 2 Plug into formula (x h) + (y k) = r (x) + (y + 2) = 32 Hint: Where is the center? How do you find it?
Suppose the equation of a circle is (x 5) + (y + 2) = 9 Write the equation of the new circle given that: A) The center of the circle moved up 4 spots and left 5: (x 0) + (y 2) = 9 Center moved from (5,-2) (0,2) B) The center of the circle moved down 3 spots and right 6: (x 11) + (y + 5) = 9 Center moved from (5,-2) (11,-5)
Find the center and radius of the circle with equation (x + 4)2 + (y 2)2 = 36. Let s Try One (x h)2 + (y k)2 = r 2 (x + 4)2 + (y 2)2 = 36 Use the standard form. Write the equation. (x ( 4))2 + (y 2)2 = 62 Rewrite the equation in standard form. h = 4 k = 2 r = 6 Find h, k, and r. The center of the circle is ( 4, 2). The radius is 6.
Graph (x 3)2 + (y + 1)2 = 4. Let s Try One (x h)2 + (y k)2 = r 2 Find the center and radius of the circle. (x 3)2 + (y ( 1))2 = 4 h = 3 k = 1 r 2 = 4, or r = 2 Draw the center (3, 1) and radius 2. Draw a smooth curve.
Equation of a Circle ( ) x h Center: Radius: r ( ) 2 2 + ( , h k ) = 2 y k r
Writing the Equation of a Circle Group x terms together, y-terms together, and move constants to the other side Complete the square for the x-terms Remember that whatever you do to one side, you must also do to the other Complete the square for the y-terms Remember that whatever you do to one side, you must also do to the other 1. 2. 3.
Example: Write the equation and find the center and radius length of : 10 8 x y x + + = 2 2 8 0 y + + = + 2 2 ( 10 ) ( x + 8 ) y + 8 x y Group terms _) ( + 8 _ _ = + + 2 2 ( 10 8 _) x x y y Complete the square 5) + y + + + = + + 2 2 ( ( 10 ) ( 5 2 4) + 8 16 ) 8 2 5 16 x x x y 49 y + + = 2 2 ( = 2 2 2 ( 5) ( ( 4 )) (7 ) x y Radius len (5, 4) Center: gth: 7
You try!!: Write the equation and find the center and radius length of : 6 12 x y x y + + + + = + = 2 2 20 0 2 2 ( 6 ) ( x 12 ) 20 x y y + + ) ( 9 + + = 0 9 2 + + 2 2 ( 6 12 36) 3 6 x x y y + + = 2 2 ( 3) ( 6) 25 x y ( 3)) + = 2 2 2 ( ( 6) (5) x y Radius len ( 3,6) Center: gth: 5
THINK ABOUT IT Find the center, the length of the radius, and write the equation of the circle if the endpoints of a diameter are (-8,2) and (2,0). Center: Use midpoint formula! with radius and an endpoint 8 2 2 0 , 2 2 Length: use distance formula + + = ( ) (2 ( 3)) + (0 1) = 26 2 2 3,1 Equation: Put it all together ) ( 3) ( x y + ( ) ( ) 2 2 ( + + = 2 2 3 ( 1) 26 x y = 2 1) 26 or
Ex. 4: Applying Graphs of Circles A bank of lights is arranged over a stage. Each light illuminates a circular area on the stage. A coordinate plane is used to arrange the lights, using the corner of the stage as the origin. The equation (x 13)2 + (y - 4)2 = 16 represents one of the disks of light. A. Graph the disk of light. B. Three actors are located as follows: Henry is at (11, 4), Jolene is at (8, 5), and Martin is at (15, 5). Which actors are in the disk of light?
Ex. 4: Applying Graphs of Circles 1. Rewrite the equation to find the center and radius. (x h)2 + (y k)2= r2 (x - 13)2 + (y - 4)2 = 16 (x 13)2 + (y 4)2= 42 The center is at (13, 4) and the radius is 4. The circle is shown on the next slide.
Ex. 4: Applying Graphs of Circles 1. Graph the disk of light The graph shows that Henry and Martin are both in the disk of light.
Ex. 4: Applying Graphs of Circles A bank of lights is arranged over a stage. Each light Use the Distance Formula Substitute values. ) 1 + ) 2 2 2 5 ( ( 1 r = r = Simplify. + ) 3 2 2 ) 4 ( 16+ 25 ( 9 r = Simplify. r = Addition Property Square root the result. r = 5
Example 3: Radio Application An amateur radio operator wants to build a radio antenna near his home without using his house as a bracing point. He uses three poles to brace the antenna. The poles are to be inserted in the ground at three points equidistant from the antenna located at J(4, 4), K( 3, 1), and L(2, 8). What are the coordinates of the base of the antenna? Step 1 Plot the three given points. Step 2 Connect J, K, and L to form a triangle.
Example 3 Continued Step 3 Find a point that is equidistant from the three points by constructing the perpendicular bisectors of two of the sides of JKL. The perpendicular bisectors of the sides of JKL intersect at a point that is equidistant from J, K, and L. The intersection of the perpendicular bisectors is P (3, 2). P is the center of the circle that passes through J, K, and L. The base of the antenna is at P (3, 2).
Check It Out! Example 3 What if ? Suppose the coordinates of the three cities in Example 3 (p. 801) are D(6, 2) , E(5, 5), and F(-2, -4). What would be the location of the weather station? Step 1 Plot the three given points. Step 2 Connect D, E, and F to form a triangle.
Check It Out! Example 3 Continued Step 3 Find a point that is equidistant from the three points by constructing the perpendicular bisectors of two of the sides of DEF. The perpendicular bisectors of the sides of DEF intersect at a point that is equidistant from D, E, and F. The intersection of the perpendicular bisectors is P(2, 1). P is the center of the circle that passes through D, E, and F. The base of the antenna is at P(2, 1).
Lesson Quiz: Part III 5. A carpenter is planning to build a circular gazebo that requires the center of the structure to be equidistant from three support columns located at E( 2, 4), F( 2, 6), and G(10, 2). What are the coordinates for the location of the center of the gazebo? (3, 1)
LOCATING THE EPICENTER OF AN EARTHQUAKE Three receiving stations are located on a coordinate plane at points (1, 4), ( 3, 1), and (5, 2). The distance from the earthquake epicenter to each station should be 2 units, 5 units, and 4 units respectively. Solution Graph the three circles. From the graph it appears that the epicenter is located at (1, 2). To check this algebraically, determine the equation for each circle and substitute x = 1 and y = 2. Example 6 2.2 - 47
LOCATING THE EPICENTER OF AN EARTHQUAKE Example 6 Station A: ( ) ) ( ( ) ) 2 2 + = x 1 y 4 4 ( 2 2 + = 1 1 2 4 4 + = 0 4 4 = 4 4 2.2 - 48
LOCATING THE EPICENTER OF AN EARTHQUAKE Example 6 Station B: ( ) ( ) 2 2 + + + = x 3 y 1 25 ( ) ( ) 2 2 + + + = 1 3 2 1 25 + = 16 9 25 = 25 25 2.2 - 49
LOCATING THE EPICENTER OF AN EARTHQUAKE Example 6 Station C: ( ) ( ) 2 2 + = x 5 y 2 16 ( ) ( ) 2 2 + = 1 5 2 2 16 + = 16 0 16 = 16 16 2.2 - 50
