What is the standard form equation for a circle? Why do you use the distance formula when writing the equation of a circle? What general equation of a.

What is the standard form equation for a circle?

Why do you use the distance formula when writing the equation of a circle?

What general equation of a circle is used when the center of the circle is

translated?How do you put an equation into standard

form?

OP = r

P(x,y)

O

r

2 2(x 0) (y 0) r

2 2x y r

2 2 2x y r

2 2 2x y r

An equation for the circle with its center at (0,0) and a radius of r is

Graph y 2 = – x 2 + 36. Identify the radius of the circle.

SOLUTION

STEP 1

Rewrite the equation y 2 = – x 2 + 36 in standard form as x 2 + y 2 = 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.

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.

Graph the equation. Identify the radius of the circle.

1. x 2 + y 2 = 9SOLUTION

STEP 1

Equation is in the standard form x 2 + y 2 = 9

STEP 2

Identify the Center and radius form the equation, the graph is a circle centered at the origin with radius

r = 9 = 3.

Draw the circle. First plot several convenient points that are 3 units from the origin, such as (0, 3), (3, 0), (0, –3), and (–3, 0). Then draw the circle that passes through the points.

2.y 2 = –x 2 + 49SOLUTION

STEP 1 Rewrite the equation y 2 = –x2 + 49 in standard form as x 2 + y 2 = 49.

STEP 2 Identify the center and radius. From the equation, the graph is a circle centered at the origin with radius

r = 49 = 7.STEP 3 Draw the circle. First plot

several convenient points that are 7 units from the origin, such as (0, 7), (7, 0), (0, –7), and (–7, 0). Then draw the circle that passes through the points.

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

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

= 29

x 2 + y 2 = r 2 Standard form

x 2 + y 2 = ( 29 )2 Substitute for r29

x 2 + y 2 = 29 Simplify

4. Write the standard form of the equation of the circle that passes through (5, –1) and whose center is the origin.

Because the point (5, –1) lies on the circle, the circle’s radius r must be the distance between the center (0, 0) and (5, –1). Use the distance formula.

SOLUTION

r = (5 – 0)2 + (–1 – 0)2 = 26 The radius is 26

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

= 26

x 2 + y 2 = r 2 Standard form

x 2 + y 2

x 2 + y 2 = 26 Simplify

SOLUTION

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

= 2 – 0 – 3 – 0

= 23

–m

23 –

the slope of the tangent line at (23, 2) is the negative reciprocal of or An equation of 3

2the tangent line is as follows:

y – 2 = (x – (– 3))32

Point-slope form32y – 2 = x +

92 Distributive property

32

13 2

y = x + Solve for y. The correct answer is C.

5. Write an equation of the line tangent to the circle x 2 + y 2 = 37 at (6, 1).

SOLUTION

A line tangent to a circle is perpendicular to the radius at the point of tangency. Because the radius

to the point (6, 1) has slopem = 1 – 0 6 – 0 =

16

the tangent line is as follows:

the slope of the tangent line at (6, 1) is the negative reciprocal of or –6 An equation of1

6

y – 1 = –6(x – 6) Point-slope form

y – 1 = –6x + 36 Distributive property

y = –6x + 37 Solve for y.

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:x 2 + y 2 < 102

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

STEP 2 Substitute the coordinates (4, 9) into the inequality from Step 1.

x 2 + y 2 < 102 Inequality from Step 1

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

The inequality is true.97 < 100So, you are in the tower’s range.

The standard equation for a translated circle is

(x – h)2 + (y – k)2 = r2

center: (h, k)

radius: r

-6-8

6

8

2

4

-4 -2

Write the standard equation of the circle graphed below.

2 2 2(x h) (y k) r 2 2 2(x ( 2)) (y 3) 4

2 2(x 2) (y 3) 16

1) C(0,0) radius: 9

Write the standard equation of a circle with the following center and radius.

2) C(2,3) radius: 53) C(-5,2) radius: 4

1) x2 + y2 = 25

Graph each equation. Label the center and radius.

2) (x – 2)2 + y2 = 43) (x + 4)2 + (y – 3)2 = 49

Write the standard equation for the circle given by x2 + y2 – 12x – 2y - 8 = 0. State the coordinates of its center and give its radius. 2 2x y 12x 2y 8 0

2 2x 12x y 2y 8 2 2(x 12x ) ( 1 136 y ) 32y 68

2 2(x 6) (y 1) 45

Center: (6,1)

Radius: 45 3 5

Write the standard equation for the circle given by x2 + y2 + 6x – 4y - 3 = 0. State the coordinates of its center and give its radius. Then sketch the graph.

2 2x y 6x 4y 3 0 2 2x 6x y 4y 3

2 2(x 6x ) ( 4 4y ) 39 4y 9 2 2(x 3) (y 2) 16

Center: (-3,2)

Radius: 16 4

-6-8

6

8

2

4

-4 -2

Write the standard equation for the circle given by x2 + y2 - 2x + 2y - 7 = 0. State the coordinates of its center and give its radius. Then sketch the graph.

Complete the square

1 1 1 1

Factor

Center is at (1,−1), r =3

p. 505, 4-18 even, 22- 28 even, 54, 56

p. 531, 4, 8, 13, 14