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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 centered in.

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Presentation on theme: "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 centered in."— Presentation transcript:

1 WARM UP Conic Sections CA ST #16

2 Standard 16 Equation of a circle with center at the origin. The equation, in standard form of the circle centered in the origin with radius r is: x 2 + y 2 = r 2. Example- x 2 + y 2 = 25 Center is (0, 0) Radius =

3 Standard 16 Graphing the example Center is (0, 0) Radius = Example- x 2 + y 2 = 25

4 Standard 16 Conic Sections CA ST #16 Equation of a circle with center at (h, k). The equation, in standard form of the circle with center (h, k) and radius r is: (x – h) 2 + (y – k) 2 = r 2. Example- (x-3) 2 + (y+2) 2 = 9 Center is (3, -2) Radius =

5 Standard 16 Graphing the example Center is (3, -2) Radius = Example- (x-3) 2 + (y+2) 2 = 9 A (3, 1`)

6 EXAMPLE 1 Graph an equation of a circle Graph y 2 = – x Identify the radius of the circle. SOLUTION STEP 1 Rewrite the equation y 2 = – x 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.

7 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.

8 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 circles 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 = The radius is 29

9 EXAMPLE 2 Write an equation of a circle 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 r 29 x 2 + y 2 = 29 Simplify

10 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

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

12 GUIDED PRACTICE for Examples 1, 2, and 3 Graph the equation. Identify the radius of the circle. 1. x 2 + y 2 = 9 SOLUTION 3

13 GUIDED PRACTICE for Examples 1, 2, and 3 2. y 2 = –x SOLUTION 7

14 GUIDED PRACTICE for Examples 1, 2, and 3 3. x 2 – 18 = –y 2 SOLUTION 18

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

16 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.

17 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.

18 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.

19 Standard 16 Classwork/ Homework Section 10-2 (page #435) From the PH book Problems 1-24 ( yes odds and evens)


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