# WARM UP Conic Sections CA ST #16.

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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 the origin with radius r is: x2 + y2 = r2 . Example- x2 + y2 = 25 Center is (0, 0) Radius =

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

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 =

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

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.

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.

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

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

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

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.

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). SOLUTION y = –6x + 37

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

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)

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