EXAMPLE 1 Graph an equation of a circle

Graph y2 = – x2 + 36. Identify the radius of the circle.

SOLUTION

STEP 1

Rewrite the equation y2 = – x2 + 36 in standard form as x2 + y2 = 36.

STEP 2

Identify the center and radius. From the equation, the graph is a circle centered at the origin with radius

r = 36 = 6.

EXAMPLE 1 Graph an equation of a circle

STEP 3

Draw the circle. First plot several convenient points that are 6 units from the origin, such as (0, 6), (6, 0), (0, –6), and (–6, 0). Then draw the circle that passes through the points.

EXAMPLE 2 Write an equation of a circle

The point (2, –5) lies on a circle whose center is the origin. Write the standard form of the equation of the circle.

SOLUTION

Because the point (2, –5) lies on the circle, the circle’s radius r must be the distance between the center (0, 0) and (2, –5). Use the distance formula.

r = (2 – 0)2 + (– 5 – 0)2

= 29= 4 + 25

The radius is 29

EXAMPLE 2 Write an equation of a circle

Use the standard form with r to write an equation of the circle.

= 29

x2 + y2 = r2 Standard form

x2 + y2 = ( 29 )2 Substitute for r29

x2 + y2 = 29 Simplify

EXAMPLE 3 Standardized Test Practice

SOLUTION

A line tangent to a circle is perpendicular to the radius at the point of tangency. Because the radius to the point (1–3, 2) has slope

= 2 – 0 – 3 – 0 = 2

3 –m

EXAMPLE 3 Standardized Test Practice

23

–the slope of the tangent line at (23, 2) is the negative reciprocal of or An equation of3

2the tangent line is as follows:

y – 2 = (x – (– 3))32

Point-slope form

32

y – 2 = x +92

Distributive property

32

13 2

y = x + Solve for y.

ANSWER

The correct answer is C.

EXAMPLE 4 Write a circular model

Cell Phones

A cellular phone tower services a 10 mile radius. You get a flat tire 4 miles east and 9 miles north of the tower. Are you in the tower’s range?

SOLUTION

STEP 1

Write an inequality for the region covered by the tower. From the diagram, this region is all points that satisfy the following inequality:

x2 + y2 < 102

In the diagram above, the origin represents the tower and the positive y-axis represents north.

EXAMPLE 4 Write a circular model

STEP 2

Substitute the coordinates (4, 9) into the inequality from Step 1.

x2 + y2 < 102 Inequality from Step 1

42 + 92 < 102? Substitute for x and y.

The inequality is true.97 < 100

ANSWER

So, you are in the tower’s range.

EXAMPLE 5 Apply a circular model

Cell Phones

In Example 4, suppose that you fix your tire and then drive south. For how many more miles will you be in range of the tower ?

SOLUTION

When you leave the tower’s range, you will be at a point on the circle x2 + y2 = 102 whose x-coordinate is 4 and whose y-coordinate is negative. Find the point (4, y) where y < 0 on the circle x2 + y2 = 102.

EXAMPLE 5 Apply a circular model

x2 + y2 = 102 Equation of the circle

42 + y2 = 102 Substitute 4 for x.

y = + 84 Solve for y.

y + 9.2 Use a calculator.

ANSWER

Because y < 0, y – 9.2. You will be in the tower’s range from (4, 9) to (4, – 9.2), a distance of | 9 – (– 9.2) | = 18.2 miles.

EXAMPLE 1 Graph an equation of an ellipse

Graph the equation 4x2 + 25y2 = 100. Identify the vertices, co-vertices, and foci of the ellipse.

SOLUTION

STEP 1

Rewrite the equation in standard form.

4x2 + 25y2 = 100 Write original equation.

4x2

100 + 25x2

100100 100= Divide each side by 100.

x2

25 +y2

4 = 1 Simplify.

EXAMPLE 1 Graph an equation of an ellipse

STEP 2

Identify the vertices, co-vertices, and foci. Note that a2 = 25 and b2 = 4, so a = 5 and b = 2. The denominator of the x2 - term is greater than that of the y2 - term, so the major axis is horizontal.

The vertices of the ellipse are at (+a, 0) = (+5, 0). The co-vertices are at (0, +b) = (0, +2). Find the foci.

c2 = a2 – b2 = 52 – 22 = 21,

so c = 21

The foci are at ( + 21 , 0), or about ( + 4.6, 0).

EXAMPLE 1 Graph an equation of an ellipse

STEP 3

Draw the ellipse that passes through each vertex and co-vertex.

EXAMPLE 2 Write an equation given a vertex and a co-vertex

Write an equation of the ellipse that has a vertex at (0, 4), a co-vertex at (– 3, 0), and center at (0, 0).

SOLUTION

Sketch the ellipse as a check for your final equation. By symmetry, the ellipse must also have a vertex at (0, – 4) and a co-vertex at (3, 0).

Because the vertex is on the y - axis and the co-vertex is on the x - axis, the major axis is vertical with a = 4, and the minor axis is horizontal with b = 3.

EXAMPLE 2 Write an equation given a vertex and a co-vertex

or

ANSWER

An equation is x2

32 +y2

42 = 1 x2 9+

y2

16= 1

EXAMPLE 3 Solve a multi-step problem

Lightning

When lightning strikes, an elliptical region where the strike most likely hit can often be identified. Suppose it is determined that there is a 50% chance that a lightning strike hit within the elliptical region shown in the diagram.

• Write an equation of the ellipse.

• The area A of an ellipse is A = π ab. Find the area of the elliptical region.

EXAMPLE 3 Solve a multi-step problem

SOLUTION

STEP 1

The major axis is horizontal, with a =400

2= 200

and b = 200 2

= 100

= 1An equation is = 1or x2

2002 + y2

1002

x2

40,000 +

y2

10,000

STEP 2

The area is A = π(200)(100) 62,800 square meters.

EXAMPLE 4 Write an equation given a vertex and a focus

Write an equation of the ellipse that has a vertex at (– 8, 0), a focus at (4, 0), and center at (0, 0).

Make a sketch of the ellipse. Because the given vertex and focus lie on the x - axis, the major axis is horizontal, with a = 8 and c = 4. To find b, use the equation c2 = a2 – b2.

SOLUTION

42 = 82 – b2

b2 = 82 – 42 = 48

EXAMPLE 4 Write an equation given a vertex and a focus

b = 48, or 34

ANSWER

An equation is x2

82 + = 1 or x2

64+

y2

48 = 1 y2

3,)2(4

4(x2 – 2x + ? ) + y2 = 8 + 4( ? )

EXAMPLE 6 Classify a conic

Classify the conic given by 4x2 + y2 – 8x – 8 = 0. Then graph the equation.

SOLUTION

Note that A = 4, B = 0, and C = 1, so the value of the discriminant is: B2 – 4AC = 02 – 4(4)(1) = – 16

Because B2– 4AC < 0 and A = C, the conic is an ellipse.To graph the ellipse, first complete the square in x.

4x2 + y2 – 8x – 8 = 0

(4x2 – 8x) + y2 = 8

4(x2 – 2x) + y2 = 8

EXAMPLE 6 Classify a conic

4(x2 – 2x + 1) + y2 = 8 + 4(1)

4(x – 1)2 + y2 = 12

(x – 1)2

3+

y2

12 = 1

From the equation, you can see that (h, k) = (1, 0), a = 12 = 2 3 , and b = 3. Use these facts to draw the ellipse.