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Transcript of 2.2 - 1 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 2 Graphs and Functions Copyright ©...
2.2 - 1Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1
2Graphs and Functions
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 2
2.2 Circles• Center-Radius Form• General Form• An Application
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 3
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 - 4Copyright © 2013, 2009, 2005 Pearson Education, Inc. 4
Center-Radius Form of the Equation of a CircleA circle with center (h, k) and radius r has equation
2 2 2,x h y k r 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
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 5
Example 1 FINDING THE CENTER-RADIUS FORM
Solution
Let (h, k) = (– 3, 4) and r = 6.
Find the center-radius form of the equation of each circle described.(a) center at (– 3, 4), radius 6
2 2 2( ) ( )x h y k r Center-radius form
2 2 2( ) ( 4)3 6x y Substitute.
Watch signs here.
2 2( 3) ( 4) 36x y Simplify.
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 6
Example 1 FINDING THE CENTER-RADIUS FORM
Solution The center is the origin and r = 3.
Find the center-radius form of the equation of each circle described.(b) center at (0, 0), radius 3
2 2 2x y r
2 2 23x y
2 2 9x y
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 7
Example 2 GRAPHING CIRCLES
Solution Writing the given equation in center-radius form
(a)
Graph each circle discussed in Example 1.2 2( 3) ( 4) 36x y
2 2 2( 3) ( 4) 6x y gives (– 3, 4) as the center and 6 as the radius.
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 8
Example 2 GRAPHING CIRCLES
Solution The graph with center (0, 0) and radius 3 is shown.
(b)
Graph each circle discussed in Example 1.2 2 9x y
2.2 - 9Copyright © 2013, 2009, 2005 Pearson Education, Inc. 9
General Form of the Equation of a CircleFor some real numbers c, d, and e, the equation
2 2 0x y cx dy e can have a graph that is a circle or a point, or is nonexistent.
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 10
General Form of the Equation of a Circle
Consider
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
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( ) ( ) .x h y k m
.m
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 11
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.
2 26 10 25 0x x y y 2 2( ) ( )6 10 25x x y y
221
( ) (6 93)2
and
221
( ) 50 2512
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 12
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.
2 2( 6 ) ( 109 25) 9 2525x x y y Add 9 and 25 on both
sides.2 2( 3) ( 5) 9x y Factor.
Complete the square.
Since 32 = 9 and 9 > 0, the equation represents a circle with center at (3, – 5) and radius 3.
2 2 2( 3) ( 5 ) 3x y
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 13
Example 4 FINDING THE CENTER AND RADIUS BY COMPLETING THE SQUARE
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 22 2 6 10 1x y x y
Divide by 2.2 2 13 5
2x y x y
2 2 13 5
2x x y y Rearrange and
regroup terms.
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 14
Example 4 FINDING THE CENTER AND RADIUS BY COMPLETING THE SQUARE
Factor and add.
2 2 13 5
225 259
44 49
4x x y y
2 23 5
92 2
x y
Center-radius form
Complete the square for both x and y.
2223 5
32 2
x y 3 5
,2 2
The equation has a circle with center atand radius 3 as its graph.
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 15
Example 5 DETERMINING WHETHER A GRAPH IS A POINT OR NONEXISTENT
The graph of the equationx2 + 10x + y2 – 4y +33 = 0
is either a point or is nonexistent. Which is it? Solution
2 210 4 33 0x x y y 2 2 3310 4x x y y Subtract 33.
2
10( 2)2
51
and2
1(
244)
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 16
Example 5 DETERMINING WHETHER A GRAPH IS A POINT OR NONEXISTENT
2 210 425 4 2533 4x x y y
Factor; add. 2 25 2 4x y
Complete the square.
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).)
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 17
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.
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 18
Example 6 LOCATING THE EPICENTER OF AN EARTHQUAKE
Station A:
2 21 4 4x y
?
2 21 21 4 4
?
0 4 4
4 4
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Example 6 LOCATING THE EPICENTER OF AN EARTHQUAKE
Station B:
2 23 1 25x y
?
2 21 23 1 25
?
16 9 25
25 25
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Example 6 LOCATING THE EPICENTER OF AN EARTHQUAKE
Station C:
2 25 2 16x y
?
2 21 25 2 16
?
16 0 16
16 16
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 21
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).