1. EXAMPLE 1
Graph an equation of a circle
Graph y2 = – x Identify the radius of the circle.
SOLUTION
STEP 1
Rewrite the equation y2 = – x 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 = = 6.

2. 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.

3. Write an equation of a circle
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 = =
The radius is 29

4. Write an equation of a circle
EXAMPLE 2
Write an equation of a circle
Use the standard form with r to write an equation of the circle. =
x2 + y2 = r2
Standard form
= ( )2
x2 + y2
Substitute for r 29
x2 + y2 = 29
Simplify

5. 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

6. Standardized Test Practice
EXAMPLE 3
Standardized Test Practice
the slope of the tangent line at (23, 2) is the negative reciprocal of or An equation of 2 3 2 – 3
the tangent line is as follows:
y – 2 = (x – (– 3)) 3 2
Point-slope form 3 2
y – 2 = x + 9
Distributive property 3 2 13
y =
x +
Solve for y.
ANSWER
The correct answer is C.

7. 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
In the diagram above, the origin represents the tower and the positive y-axis represents north.
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

8. Substitute the coordinates (4, 9) into the inequality from Step 1.
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
< 102 ?
Substitute for x and y.
97 < 100 
The inequality is true.
ANSWER
So, you are in the tower’s range.

9. 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.

10. EXAMPLE 5 Apply a circular model x2 + y2 = 102 42 + y2 = 102 84 + y =
Equation of the circle
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) | = miles.

11. Graph an equation of an ellipse
EXAMPLE 1
Graph an equation of an ellipse
Graph the equation 4x y2 = Identify the vertices, co-vertices, and foci of the ellipse.
SOLUTION
STEP 1
Rewrite the equation in standard form.
4x y2 = 100
Write original equation.
4x2
100 +
25x2 =
Divide each side by 100. x2 25 +
y24
= 1
Simplify.

12. 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 – = 21,
so c = 21
The foci are at ( , 0), or about ( + 4.6, 0).

13. EXAMPLE 1
Graph an equation of an ellipse
STEP 3
Draw the ellipse that passes through each vertex and co-vertex.

14. 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.

15. EXAMPLE 2
Write an equation given a vertex and a co-vertex
ANSWER
x2 32 +
y2 42
x2 9 +
y2 16
= 1
An equation is
= 1 or

16. 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.

17. 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
= 1
An equation is
= 1 or
x2 2002 +
y2 1002
x2 40,000
y2 10,000
STEP 2
The area is A = π(200)(100) ,800 square meters.

18. 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).
SOLUTION
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.
42 = 82 – b2
b2 = 82 – 42 = 48

19. EXAMPLE 4
Write an equation given a vertex and a focus
b =
48, or 3 4
ANSWER
An equation is
x2 82 +
= 1 or
x2 64
y2 48 y2
3,)2 (4

20. 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
4(x2 – 2x + ? ) + y2 = 8 + 4( ? )

21. 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 = , and b = 3. Use these facts to draw the ellipse.