WARM UP Conic Sections CA ST #16

Standard 16 Equation of a circle with center at the origin.

The equation, in standard form of the circle centeredin the origin with radius r is:

x2 + y2 = r2 .

Example- x2 + y2 = 25

Center is (0, 0)Radius = 25 5

Standard 16 Graphing the example

Center is (0, 0)Radius = 25 5

Example- x2 + y2 = 25

Standard 16 Conic Sections CA ST #16

Equation of a circle with center at (h, k).

The equation, in standard form of the circle withcenter (h, k) and radius r is:

(x – h)2 + (y – k)2 = r2 .

Example- (x-3)2 + (y+2)2 = 9

Center is (3, -2)Radius = 9 3

Standard 16 Graphing the example

Center is (3, -2)Radius = 9 3

Example- (x-3)2 + (y+2)2 = 9

A (3, 1`)

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 (–3, 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.

GUIDED PRACTICE for Examples 1, 2, and 3

Graph the equation. Identify the radius of the circle.

1. x2 + y2 = 9

SOLUTION 3

GUIDED PRACTICE for Examples 1, 2, and 3

2. y2 = –x2 + 49

SOLUTION 7

GUIDED PRACTICE for Examples 1, 2, and 3

3. x2 – 18 = –y2

SOLUTION 18

GUIDED PRACTICE for Examples 1, 2, and 3

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

SOLUTION x2 + y2 = 26

5. Write an equation of the line tangent to the circle x2 + y2 = 37 at (6, 1).

y = –6x + 37SOLUTION

EXAMPLE 1 Graph the equation of circle with center (h. k)

Graph (x – 2)2 + (y + 3) 2 = 9.

SOLUTION

STEP 1

Compare the given equation to the standard form of an equation of a circle. You can see that the graph is a circle with center at (h, k) = (2, –3) and radius r = 9 = 3.

EXAMPLE 1 Graph the equation of a translated circle

STEP 2

Plot the center. Then plot several points that are each 3 units from the center:

(2 + 3, –3) = (5, –3) (2 – 3, –3) = (–1, –3)

(2, –3 + 3) = (2, 0) (2, –3 – 3) = (2, –6)

STEP 3

Draw a circle through the points.

GUIDED PRACTICE for Examples 1 and 2

Graph (x + 1)2 + (y – 3) 2 = 4.

SOLUTION

circle with center at (h, k) = (– 1, 3) and radius r = 2

1.

Standard 16 Classwork/ Homework

Section 10-2 (page #435)From the PH bookProblems 1-24 ( yes odds and evens)