1. EXAMPLE 1 Graph an equation of a circle 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.

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. 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

4. 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

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. EXAMPLE 3 Standardized Test Practice 2 3 – the slope of the tangent line at (23, 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 + 9 2 Distributive property 3 2 13 2 y = x + Solve for y. ANSWER The correct answer is C.

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

8. GUIDED PRACTICE for Examples 1, 2, and 3 STEP 2 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.

9. GUIDED PRACTICE for Examples 1, 2, and 3 2. y 2 = –x 2 + 49 SOLUTION 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.

10. GUIDED PRACTICE for Examples 1, 2, and 3 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.

11. GUIDED PRACTICE for Examples 1, 2, and 3 STEP 2 Identify the center and radius. From the equation, the graph is a circle centered at the origin with radius 3. x 2 – 18 = –y 2 SOLUTION STEP 1 Rewrite the equation x 2 – 18 = –y 2 in standard form as x 2 + y 2 = 18. r = 18 = 3 2.

12. GUIDED PRACTICE for Examples 1, 2, and 3 STEP 3 Draw the circle. First plot several convenient points that are 3 2 units from the origin, such as (0, 3 2 ), (3 2, 0), (0, –3 2 ), and (–3 2, 0). Then draw the circle that passes through the points.

13. 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. 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

14. GUIDED PRACTICE for Examples 1, 2, and 3 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 = ( 26 ) 2 Substitute for r 26 x 2 + y 2 = 26 Simplify

15. GUIDED PRACTICE for Examples 1, 2, and 3 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 slope m = 1 – 0 6 – 0 = 1 6 the tangent line is as follows: the slope of the tangent line at (6, 1) is the negative reciprocal of or –6 An equation of 1 6 y – 1 = –6(x – 6) Point-slope form

16. GUIDED PRACTICE for Examples 1, 2, and 3 y – 1 = –6x + 36 Distributive property y = –6x + 37 Solve for y.