Transcript Circles

2
Graphs and
Functions
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
12.2 - 1
2.2
Circles
• Center-Radius Form
• General Form
• An Application
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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.
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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
2
2
2
x y r .
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42.2 - 4
FINDING THE CENTER-RADIUS
FORM
Example 1
Find the center-radius form of the equation of
each circle described.
(a) center at (–3, 4), radius 6
Solution
(x  h)  (y  k )  r
2
 (  3 )  ( y  4 )  6
2
2
x
2
2
2
Watch
signs here.
Center-radius form
Substitute.
Let (h, k) = (– 3, 4) and r = 6.
( x  3)  (y  4)  36
2
2
Simplify.
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Example 1
FINDING THE CENTER-RADIUS
FORM
Find the center-radius form of the equation of
each circle described.
(b) center at (0, 0), radius 3
Solution
The center is the origin and r = 3.
x y
2
r
2
x y
2
3
2
x y
2
9
2
2
2
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Example 2
GRAPHING CIRCLES
Graph each circle discussed in Example 1.
(a) ( x  3 ) 2  ( y  4 ) 2  3 6
Solution Writing the given equation in
center-radius form
 x  (  3 )  ( y  4 )  6
2
2
2
gives (–3, 4) as the
center and 6 as the
radius.
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Example 2
GRAPHING CIRCLES
Graph each circle discussed in Example 1.
(b) x 2  y 2  9
Solution The
graph with center
(0, 0) and radius 3
is shown.
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General Form of the Equation
of a Circle
For some real numbers c, d, and e, the
equation
x  y  cx  d y  e  0
2
2
can have a graph that is a circle or a
point, or is nonexistent.
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92.2 - 9
General Form of the Equation of a
Circle
Consider ( x  h ) 2  ( y  k ) 2  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 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.
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Example 3
FINDING THE CENTER AND RADIUS BY
COMPLETING THE SQUARE
Show that x2 – 6x + y2 +10y + 25 = 0 has a
circle as its 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
)  (y  10 y
2
2
2
1

2
(6 )  (3 )  9
 2

)  25
2
and
1

2
(1 0 )  5  2 5
 2

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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.
( x  6 x  9 )  ( y  10 y  25 )  25  9  25
2
Add 9 and
25 on both
sides.
2
Complete the square.
(x  3)  (y  5)  9
2
2
( x  3 )  ( y   5 )  3
2
2
Factor.
2
Since 32 = 9 and 9 > 0, the equation
represents a circle with center at (3, –5) and
radius 3.
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FINDING THE CENTER AND RADIUS BY
COMPLETING THE SQUARE
Example 4
Show that 2x2 + 2y2 – 6x +10y = 1 has a circle
as its graph. Find the center and radius.
Solution To complete the square, the
coefficients of the x2- and y2-terms must be 1.
2 x  2y  6 x  10 y  1
2
2
x  y  3x  5y 
2
x
2
 3x
2
  y
2
 5y

1
Divide by 2.
2
1
2
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Rearrange and
regroup terms.
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FINDING THE CENTER AND RADIUS BY
COMPLETING THE SQUARE
Example 4
9  2
25
1 9 25
 2
x  3x 
 y  5y 
  

4 
4 
2 4
4
2
3
5


x 
 y 


2
2
2
3


 5 
x 
 y  






2
2 
Complete the square
for both x and y.
2
 9
Factor and add.
2
 3
2
Center-radius form
The equation has a circle with center at
and radius 3 as its graph.
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5
3
,
2
2
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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
x  10 x  y  4 y  33  0
2
2
x  10 x  y  4 y  33
2
2
2
1

(1 0 )  2 5
 2

Subtract 33.
2
and
1

( 4)  4
 2

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Example 5
DETERMINING WHETHER A GRAPH IS A
POINT OR NONEXISTENT
 x  10 x  25    y  4 y  4   33  25  4
2
2
Complete the square.
x  5  y  2   4
2
2
Factor; add.
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).)
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Example 6
LOCATING THE EPICENTER OF
AN EARTHQUAKE
Suppose receiving stations A, B, and C are
located on a coordinate plane at the points
(1, 4), (– 3, –1), and (5, 2). Let the distances from
the earthquake epicenter to these stations be 2
units, 5 units, and 4 units, respectively. Where on
the coordinate plane is the epicenter located?
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.
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Example 6
LOCATING THE EPICENTER OF
AN EARTHQUAKE
Station A:
 x  1   y  4   4
2
1  1
2
2
?
 2  4  4
2
?
0  44
44
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Example 6
LOCATING THE EPICENTER OF
AN EARTHQUAKE
Station B:
 x  3    y  1  2 5
2
1  3 
2
2
?
  2  1  2 5
2
?
16  9  25
25  25
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Example 6
LOCATING THE EPICENTER OF
AN EARTHQUAKE
Station C:
 x  5    y  2   16
2
1  5 
2
2
?
 2  2  16
2
?
16  0  16
16  16
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Example 6
LOCATING THE EPICENTER OF
AN EARTHQUAKE
The point (1, 2) lies on
all three graphs.
Thus, we can
conclude that the
epicenter is at (1, 2).
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