2Standard 16Equation 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 = 25Center is (0, 0)Radius =
3Standard 16Graphing the exampleExample- x2 + y2 = 25Center is (0, 0)Radius =
4Standard 16Conic Sections CA ST #16Equation 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 = 9Center is (3, -2)Radius =
5Standard 16 Graphing the example Example- (x-3)2 + (y+2)2 = 9 Center is (3, -2)Radius =
6EXAMPLE 1Graph an equation of a circleGraph y2 = – x Identify the radius of the circle.SOLUTIONSTEP 1Rewrite the equation y2 = – x in standard form as x2 + y2 = 36.STEP 2Identify the center and radius. From the equation, the graph is a circle centered at the origin with radiusr = = 6.
7EXAMPLE 1Graph an equation of a circleSTEP 3Draw 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.
8Write an equation of a circle EXAMPLE 2Write an equation of a circleThe point (2, –5) lies on a circle whose center is the origin. Write the standard form of the equation of the circle.SOLUTIONBecause 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 is29
9Write an equation of a circle EXAMPLE 2Write an equation of a circleUse the standard form with r to write an equation of the circle.=x2 + y2 = r2Standard form= ( )2x2 + y2Substitute for r29x2 + y2 = 29Simplify
10EXAMPLE 3Standardized Test PracticeSOLUTIONA 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=23–m
11Standardized Test Practice EXAMPLE 3Standardized Test Practicethe slope of the tangent line at (–3, 2) is the negative reciprocal of or An equation of232–3the tangent line is as follows:y – 2 = (x – (– 3))32Point-slope form32y – 2 = x +9Distributive property3213y =x +Solve for y.ANSWERThe correct answer is C.
12GUIDED PRACTICEfor Examples 1, 2, and 3Graph the equation. Identify the radius of the circle.1.x2 + y2 = 9SOLUTION3
15GUIDED PRACTICEfor Examples 1, 2, and 34. Write the standard form of the equation of the circle that passes through (5, –1) and whose center is the origin.SOLUTIONx2 + y2 = 265. Write an equation of the line tangent to the circle x2 + y2 = 37 at (6, 1).SOLUTIONy = –6x + 37
16EXAMPLE 1Graph the equation of circle with center (h. k)Graph (x – 2)2 + (y + 3) 2 = 9.SOLUTIONSTEP 1Compare 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.
17EXAMPLE 1Graph the equation of a translated circleSTEP 2Plot 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 3Draw a circle through the points.
18GUIDED PRACTICEfor Examples 1 and 21.Graph (x + 1)2 + (y – 3) 2 = 4.SOLUTIONcircle with center at (h, k) = (– 1, 3) and radius r = 2
19Problems 1-24 ( yes odds and evens) Standard 16Classwork/ HomeworkSection 10-2 (page #435)From the PH bookProblems 1-24 ( yes odds and evens)