Transcript Circles - SaigonTech

10TH
EDITION
COLLEGE
ALGEBRA
LIAL
HORNSBY
SCHNEIDER
2.2 - 1
2.2 Circles
Center-Radius Form
General Form
An Application
2.2 - 2
Circle-Radius form
By definition, a circle is the set of all points
in a plane that lie a given distance from a
given point. The given distance is the
radius of the circle, and the given point is
the center.
2.2 - 3
Center-Radius Form of the
Equation of a Circle
A circle with center (h, k) and radius r has
equation
 x  h   y  k   r ,
2
2
2
which is the center-radius form of the
equation of the circle. A circle with center
(0, 0) and radius r has equation
x y r .
2
2
2
2.2 - 4
FINDING THE CENTER-RADIUS
FORM
Example 1
Find the center-radius form of a circle with a
center at (–3, 4), radius 6.
Solution:
a. Use (h, k) = (–3, 4) and r = 6
( x  h)  ( y  k )  r
2
2
2
2
2
x

(

3
)

(
y

4)

6


2
Center-radius form
Substitute
Watch
signs here.
2.2 - 5
Example 1
FINDING THE CENTER-RADIUS
FORM
Find the center-radius form of a circle with a
center at (–3, 4), radius 6.
a. Use (h, k) = (–3, 4) and r = 6
Solution
2
2
2
 x  ( 3)  ( y  4)  6
Substitute
( x  3)2  ( y  4)2  36
2.2 - 6
Example 1
FINDING THE CENTER-RADIUS
FORM
b. Find the center-radius form of a circle with
a center at (0, 0), radius 3.
Solution
Because the center is the origin and r = 3, the
equation is
x2  y 2  r 2
x y 3
2
2
2
x y 9
2
2
2.2 - 7
Example 2
GRAPHING CIRCLES
Graph the circle.
a. ( x  3)  ( y  4)  36
2
2
Solution
 x  ( 3)
2
 ( y  4)  6
2
2
Gives (–3, 4) as the center and 6 as the
radius.
2.2 - 8
Example 2
GRAPHING CIRCLES
y
Graph the circle.
a. ( x  3)  ( y  4)  36
2
Solution
2
(x, y)
6
(–3, 4)
x
Gives (–3, 4) as the center
and 6 as the radius.
2.2 - 9
Example 2
GRAPHING CIRCLES
y
Graph the circle.
b. x  y  9
2
2
Solution
3
(x, y)
x
Gives (0, 0) as the center and
3 as the radius.
2.2 - 10
General Form of the Equation
of a Circle
The equation
x  y  cx  dy  e  0
2
2
for some real numbers c, d, and e, can
have a graph that is a circle or a point,
or is nonexistent.
2.2 - 11
General Form of the Equation of a
Circle
Consider ( x  h )  ( y  k )  m
There are three possibilities for the graph
based on the value of m.
1. If m > 0, then r 2 = m, and the graph of the
equation is a circle with the radius m.
2. If m = 0, then the graph of the equation is
the single point (h, k).
3. If m < 0, then no points satisfy the
equation and the graph is nonexistent.
2
2
2.2 - 12
Example 3
FINDING THE CENTER AND RADIUS BY
COMPLETING THE SQUARE
Show that x2 – 6x + y2 +10y + 25 = 0 has a
circle as a graph. Find the center and radius.
Solution We complete the square twice,
once for x and once for y.
x  6 x  y  10 y  25  0
2
2
( x  6x
2
2
)  ( y  10 y
2
)  25
2
1

1

2
2
(

6
)

(

3)

9
(
1
0
)

5

25
and
 2

 2

2.2 - 13
Example 3
FINDING THE CENTER AND RADIUS BY
COMPLETING THE SQUARE
Add 9 and 25 on the left to complete the two
squares, and to compensate, add 9 and 25
on the right.
Complete the square.
( x  6 x  9)  ( y  10 y  25)  25  9  25
2
Add 9 and
25 on both
sides.
2
( x  3)2  ( y  5)2  9
Factor
Since 9 > 0, the equation represents a circle
with center at (3, –5) and radius 3.
2.2 - 14
Example 4
FINDING THE CENTER AND RADIUS BY
COMPLETING THE SQUARE
Show that 2x2 + 2y2 – 6x +10y = 1 has a circle
as a graph. Find the center and radius.
Solution To complete the square, the
coefficients of the x2- and y2-terms must be 1.
2 x 2  2 y 2  6 x  10 y  1
2  x 2  3 x   2  y 2  5y   1
Group the terms; factor
out 2.
2.2 - 15
Example 4
FINDING THE CENTER AND RADIUS BY
COMPLETING THE SQUARE
2  x  3 x   2  y  5y   1
2
2
Group the terms; factor
out 2.
9
25 
 2
 2
2  x  3 x    2  y  5y  
4
4 


9
 25 
 1 2    2  
Be careful
here.
4
 4 
2.2 - 16
FINDING THE CENTER AND RADIUS BY
COMPLETING THE SQUARE
Example 4
9
25 
 2
 2
2  x  3 x    2  y  5y  
4
4 


9
 25 
 1 2    2  
4
 4 
2
2
3
5


2  x    2  y    18
2
2


2
Factor; simplify on
the right.
2
3 
5

x   y    9
2 
2

Divide both sides
by 2.
2.2 - 17
FINDING THE CENTER AND RADIUS BY
COMPLETING THE SQUARE
Example 4
2
2
3 
5

x   y    9
2 
2

Divide both sides
by 2.
3 5
The equation has a circle with center at  ,  
2 2
and radius 3 as its graph.
2.2 - 18
DETERMINING WHETHER A GRAPH IS A
POINT OR NONEXISTENT
Example 5
The graph of the equation
x2 + 10x + y2 – 4y +33 = 0
is either a point or is nonexistent. Which is it?
Solution We complete the square for x and y.
x  10 x  y  4 y  33  0
2
2
x  10 x  y  4 y  33
2
2
2
1

(
10
)

2
5
 2

Subtract 33.
2
and
1

(

4)

4
 2

2.2 - 19
Example 5
DETERMINING WHETHER A GRAPH IS A
POINT OR NONEXISTENT
The graph of the equation
x2 + 10x + y2 – 4y +33 = 0
is either a point or is nonexistent. Which is it?
2
2
1
1



and
(
10
)

2
5
(

4)

4
 2

 2

Complete the square.
x
2
 10 x  25    y  4 y  4   33  25  4
2
 x  5    y  2   4
2
2
Factor; add.
2.2 - 20
Example 5
DETERMINING WHETHER A GRAPH IS A
POINT OR NONEXISTENT
Since –4 < 0, there are no ordered pairs (x, y),
with both x and y both real numbers, satisfying
the equation. The graph of the given equation
is nonexistent; it contains no points. ( If the
constant on the right side were 0, the graph
would consist of the single point (–5, 2).)
2.2 - 21
Example 6
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.
2.2 - 22
Example 6
LOCATING THE EPICENTER OF
AN EARTHQUAKE
Station A:
 x  1   y  4   4
2
2
1  1   2  4   4
2
2
04 4
44
2.2 - 23
Example 6
LOCATING THE EPICENTER OF
AN EARTHQUAKE
Station B:
 x  3    y  1  25
2
2
1  3    2  1  25
2
2
16  9  25
25  25
2.2 - 24
Example 6
LOCATING THE EPICENTER OF
AN EARTHQUAKE
Station C:
 x  5    y  2   16
2
2
1  5    2  2   16
2
2
16  0  16
16  16
2.2 - 25
Example 6
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.
The point (1, 2) does lie on all three graphs;
thus, we can conclude that the epicenter is
at (1, 2).
2.2 - 26