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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 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)