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

6. EXAMPLE 1
Graph an equation of a circle
Graph y2 = – x Identify the radius of the circle.
SOLUTION
STEP 1
Rewrite the equation y2 = – x 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 = = 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. Write an equation of a circle
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 = =
The radius is 29

9. Write an equation of a circle
EXAMPLE 2
Write an equation of a circle
Use the standard form with r to write an equation of the circle. =
x2 + y2 = r2
Standard form
= ( )2
x2 + y2
Substitute for r 29
x2 + y2 = 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. Standardized Test Practice
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 2 – 3
the tangent line is as follows:
y – 2 = (x – (– 3)) 3 2
Point-slope form 3 2
y – 2 = x + 9
Distributive property 3 2 13
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.
x2 + y2 = 9
SOLUTION 3

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

14. GUIDED PRACTICE
for Examples 1, 2, and 3 3.
x2 – 18 = –y2
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
x2 + y2 = 26
5. Write an equation of the line tangent to the circle x2 + y2 = 37 at (6, 1).
SOLUTION
y = –6x + 37

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 1.
Graph (x + 1)2 + (y – 3) 2 = 4.
SOLUTION
circle with center at (h, k) = (– 1, 3) and radius r = 2

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