44
CHAPTER 1
1.5
Graphs
Circles
OBJECTIVES
1
2
3
Write the Standard Form of the Equation of a Circle
Graph a Circle by Hand and by Using a Graphing Utility
Work with the General Form of the Equation of a Circle
1 Write the Standard Form of the Equation of a Circle
✓
One advantage of a coordinate system is that it enables us to translate a geometric
statement into an algebraic statement, and vice versa. Consider, for example, the following geometric statement that defines a circle.
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SECTION 1.5
Circles
45
A circle is a set of points in the xy-plane that are a fixed distance r from a fixed
point 1h, k2. The fixed distance r is called the radius, and the fixed point 1h, k2
is called the center of the circle.
Figure 61 shows the graph of a circle. To find the equation, we let 1x, y2 represent the coordinates of any point on a circle with radius r and center 1h, k2. Then the
distance between the points 1x, y2 and 1h, k2 must always equal r. That is, by the distance formula
Figure 61
y
(x, y )
r
2
2
41x - h2 + 1y - k2 = r
(h, k )
x
or, equivalently,
1x - h22 + 1y - k22 = r2
The standard form of an equation of a circle with radius r and center 1h, k2 is
1x - h22 + 1y - k22 = r2
(1)
The standard form of an equation of a circle of radius r with center at the origin 10, 02 is
x2 + y2 = r2
If the radius r = 1, the circle whose center is at the origin is called the unit
circle and has the equation
x2 + y2 = 1
See Figure 62.
Figure 62
Unit circle x2 + y2 = 1
y
2
2 Y1 1 x
3
1
3
2 Y 1 x 2
2
1
(0,0)
1
1
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x
46
CHAPTER 1
Graphs
EXAMPLE 1
Writing the Standard Form of the Equation of a Circle
Write the standard form of the equation of the circle with radius 5 and center
1-3, 62.
Solution
Using the form of equation (1) and substituting the values r = 5, h = -3, and
k = 6, we have
1x - h22 + 1y - k22 = r2
1x + 322 + 1y - 622 = 25
NOW WORK PROBLEM
5.
2 Graph a Circle by Hand and by Using a Graphing Utility
✓
The graph of any equation of the form (1) is that of a circle with radius r and center
1h, k2.
EXAMPLE 2
Graphing a Circle by Hand and by Using a Graphing Utility
Graph the equation:
Solution
1x + 322 + 1y - 222 = 16
The graph of the equation is a circle. To graph the equation by hand, we first compare the given equation to the standard form of the equation of a circle. The comparison yields information about the circle.
1x + 322 + 1y - 222 = 16
1x - 1-3222 + 1y - 222 = 4 2
q
q
q
1x - h2 + 1y - k2 = r2
2
We see that h = -3, k = 2, and r = 4. The circle has center 1-3, 22 and a radius of
4 units. To graph this circle, we first plot the center 1-3, 22. Since the radius is 4, we
can locate four points on the circle by plotting points 4 units to the left, to the right,
up, and down from the center. These four points can then be used as guides to obtain
the graph. See Figure 63.
To graph a circle on a graphing utility, we must write the equation in the form
y = 5expression involving x6.* We must solve for y in the equation
Figure 63
(–3, 6)
y
6
1x + 322 + 1y - 222 = 16
4
(–7, 2)
–10
(–3, 2)
1y - 222 = 16 - 1x + 322
(1, 2)
y - 2 = ; 416 - 1x + 322
2 x
–5
2
y = 2 ; 416 - 1x + 322
(–3, –2)
Subtract (x + 3)2 from both sides.
Use the Square Root Method.
Add 2 to both sides.
*
Some graphing utilities (e.g., TI-83, TI-84, and TI-86) have a CIRCLE function that allows the user to
enter only the coordinates of the center of the circle and its radius to graph the circle.
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SECTION 1.5
Circles
47
To graph the circle, we graph the top half
Figure 64
Y1 = 2 + 416 - 1x + 322
6
Y1
and the bottom half
–9
Y2 = 2 - 416 - 1x + 322
Also, be sure to use a square screen. Otherwise, the circle will appear distorted.
Figure 64 shows the graph on a TI-84 Plus.
3
Y2
–2
NOW WORK PROBLEMS
EXAMPLE 3
21(a)
AND
(b).
Finding the Intercepts of a Circle
For the circle 1x + 322 + 1y - 222 = 16, find the intercepts, if any, of its graph.
Solution
This is the equation discussed and graphed in Example 2. To find the x-intercepts, if
any, let y = 0 and solve for x. Then
1x + 322 + 1y - 222 = 16
1x + 322 + 10 - 222 = 16
y = 0
1x + 32 + 4 = 16
2
Simplify.
1x + 32 = 12
2
Simplify.
x + 3 = ; 212
x = -3 ; 223
Apply the Square Root Method.
Solve for x.
The x-intercepts are -3 - 223 L -6.46 and -3 + 223 L 0.46.
To find the y-intercepts, if any, we let x = 0 and solve for y. Then
1x + 322 + 1y - 222 = 16
10 + 322 + 1y - 222 = 16
x = 0
9 + 1y - 22 = 16
2
Simplify.
1y - 22 = 7
2
Simplify.
y - 2 = ; 27
y = 2 ; 27
Apply the Square Root Method.
Solve for y.
The y-intercepts are 2 - 27 L -0.65 and 2 + 27 L 4.65.
Look back at Figure 63 to verify the approximate locations of the intercepts. NOW WORK PROBLEM
21(C).
3 Work with the General Form of the Equation of a Circle
✓
If we eliminate the parentheses from the standard form of the equation of the circle
given in Example 3, we get
1x + 322 + 1y - 222 = 16
x2 + 6x + 9 + y2 - 4y + 4 = 16
which we find, upon simplifying, is equivalent to
x2 + y2 + 6x - 4y - 3 = 0
It can be shown that any equation of the form
x2 + y2 + ax + by + c = 0
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48
CHAPTER 1
Graphs
has a graph that is a circle, or a point, or has no graph at all. For example, the graph
of the equation x2 + y2 = 0 is the single point 10, 02. The equation
x2 + y2 + 5 = 0, or x2 + y2 = -5, has no graph, because sums of squares of real
numbers are never negative. When its graph is a circle, the equation
x2 + y2 + ax + by + c = 0
is referred to as the general form of the equation of a circle.
If an equation of a circle is in the general form, we use the method of completing the square to put the equation in standard form so that we can identify its center
and radius.
EXAMPLE 4
Graphing a Circle Whose Equation Is in General Form
Graph the equation x2 + y2 + 4x - 6y + 12 = 0
Solution
We complete the square in both x and y to put the equation in standard form. Group
the expression involving x, group the expression involving y, and put the constant on
the right side of the equation. The result is
1x2 + 4x2 + 1y2 - 6y2 = -12
Next, complete the square of each expression in parentheses. Remember that any
number added on the left side of the equation must be added on the right.
(x2 4x 4) (y2 6y 9) 12 4 9
( 42 )
2
(6
)
2
2
4
9
(x 2)2 (y 3)2 1
Factor.
We recognize this equation as the standard form of the equation of a circle with radius 1 and center 1-2, 32. To graph the equation by hand, use the center 1-2, 32 and
the radius 1. See Figure 65 (a).
To graph the equation using a graphing utility, we need to solve for y.
1y - 322 = 1 - 1x + 222
y - 3 = ; 41 - 1x + 222
Use the Square Root Method.
y = 3 ; 41 - 1x + 222
Add 3 to both sides.
Figure 65(b) illustrates the graph.
Figure 65
y
(2, 4)
4
1
(3, 3)
(1, 3)
Y1 3 1 (x 2)2 5
(2, 3)
(2, 2)
3
1 x
4.5
1.5
Y2 3 1 (x 2)2
1
(b)
(a)
NOW WORK PROBLEM
25.
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SECTION 1.5
EXAMPLE 5
Circles
49
Finding the General Equation of a Circle
Find the general equation of the circle whose center is 11, -22 and whose graph
contains the point 14, -22.
Solution
Figure 66
y
3
r
(1, 2)
5
x
To find the equation of a circle, we need to know its center and its radius. Here, we
know that the center is 11, -22. Since the point 14, -22 is on the graph, the radius r
will equal the distance from 14, -22 to the center 11, -22. See Figure 66. Thus,
r = 414 - 122 + 3-2 - 1-2242
= 29 = 3
The standard form of the equation of the circle is
1x - 122 + 1y + 222 = 9
(4, 2)
Eliminating the parentheses and rearranging terms, we get the general equation
5
x2 + y2 - 2x + 4y - 4 = 0
Overview
The discussion in Sections 1.4 and 1.5 about circles and lines dealt with two main
types of problems that can be generalized as follows:
1. Given an equation, classify it and graph it.
2. Given a graph, or information about a graph, find its equation.
This text deals with both types of problems. We shall study various equations,
classify them, and graph them. Although the second type of problem is usually more
difficult to solve than the first, in many instances a graphing utility can be used to
solve such problems.
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