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WARM UP

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 the origin with radius r is: x 2 + y 2 = r 2. Example- x 2 + y 2 = 25. Center is (0, 0) Radius =. Standard 16.

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WARM UP

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  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: x2 + y2 = r2. Example- x2 + y2 = 25 Center is (0, 0) Radius =

  3. Standard 16 Graphing the example Example- x2 + y2 = 25 Center is (0, 0) Radius =

  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 = r2. Example- (x-3)2 + (y+2)2 = 9 Center is (3, -2) Radius =

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

  6. 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 Graphy2 = – x2 + 36. Identify the radius of the circle. SOLUTION STEP 1 Rewrite the equationy2 = – x2 + 36in standard form asx2 + y2 = 36. STEP 2

  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. The radius is = 29 r = (2 – 0)2 + (–5 – 0)2 = 4 + 25 29 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 rmust be the distance between the center (0, 0) and (2, –5). Use the distance formula.

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

  10. 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 2 – 3 – 0 3 EXAMPLE 3 Standardized Test Practice SOLUTION m

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

  12. for Examples 1, 2, and 3 GUIDED PRACTICE Graph the equation. Identify the radius of the circle. 1. x2 + y2 = 9 3 SOLUTION

  13. 2. y2 = –x2 + 49 for Examples 1, 2, and 3 GUIDED PRACTICE SOLUTION 7

  14. for Examples 1, 2, and 3 GUIDED PRACTICE 3. x2 – 18 = –y2 SOLUTION 18

  15. for Examples 1, 2, and 3 GUIDED PRACTICE 4. Write the standard form of the equation of the circle that passes through (5, –1) and whose center is the origin. x2 + y2 = 26 SOLUTION 5. Write an equation of the line tangent to the circle x2 + y2=37 at (6, 1). SOLUTION y = –6x + 37

  16. 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= = 3. 9 EXAMPLE 1 Graph the equation of circle with center (h. k) Graph(x – 2)2 + (y + 3) 2 = 9. SOLUTION STEP 1

  17. EXAMPLE 1 Graph the equation of a translated circle STEP 2 Plot the center. Then plot several points that are each 3units 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. for Examples 1 and 2 GUIDED PRACTICE 1. Graph(x + 1)2 + (y – 3) 2 = 4. SOLUTION circle with center at (h, k) = (– 1, 3)and radius r=2

  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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