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


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.


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


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.


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


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.


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


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.


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


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



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



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.



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.


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)


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).)


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.



Station A:

2 21 4 4x y

?

2 21 21 4 4

?

0 4 4

4 4



Station B:

2 23 1 25x y

?

2 21 23 1 25

?

16 9 25

25 25



Station C:

2 25 2 16x y

?

2 21 25 2 16

?

16 0 16

16 16



The point (1, 2) lies on all three graphs. Thus, we can conclude that the epicenter is at (1, 2).

2.2 - 1 copyright © 2013, 2009, 2005 pearson education, inc. 1 2 graphs and functions copyright ©...

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