PowerPoint Presentation : Introduction A Closer Look at Graphing Translations Definition of a Circle Definition of Radius Definition of a Unit Circle Deriving the Equation of a Circle The Distance Formula Moving The Center Completing the Square It is a Circle if? Conic Movie Conic Collage End Next Geometers Sketchpad: Cosmos Geometers Sketchpad: Headlights Projectile Animation Planet Animation Plane Intersecting a Cone Animation Footnotes Skip Intro
PowerPoint Presentation : Next CONIC SECTIONS Parabola Circle Ellipse Hyperbola Quadratic Relations Previous Main Menu End
PowerPoint Presentation : The quadratic relations that we studied in the beginning of the year were in the form of y = Ax 2 + Dx + F , where A, D, and F stand for constants, and A ≠ 0. This quadratic relation is a parabola, and it is the only one that can be a function. It does not have to be a function, though. A parabola is determined by a plane intersecting a cone and is therefore considered a conic section. Next Previous Main Menu End
PowerPoint Presentation : The general equation for all conic sections is: Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 where A, B, C, D, E and F represent constants When an equation has a “ y 2 ” term and/or an “ xy ” term it is a quadratic relation instead of a quadratic function. and where the equal sign could be replaced by an inequality sign. Next Previous Main Menu End
PowerPoint Presentation : A Closer Look at Graphing Conics Plot the graph of each relation. Select values of x and calculate the corresponding values of y until there are enough points to draw a smooth curve. Approximate all radicals to the nearest tenth. x 2 + y 2 = 25 x 2 + y 2 + 6x = 16 x 2 + y 2 - 4y = 21 x 2 + y 2 + 6x - 4y = 12 Conclusions Next Previous Main Menu Last Viewed End
PowerPoint Presentation : x 2 + y 2 = 25 x = 3 3 2 + y 2 = 25 9 + y 2 = 25 y 2 = 16 There are 2 points to graph: (3,4) (3,-4) y = ± 4 Next Previous Main Menu A Close Look at Graphing End
PowerPoint Presentation : Continue to solve in this manner, generating a table of values. -5 0 -4 ±3 -3 ±4 -2 ±4.6 -1 ±4.9 0 ±5 1 ±4.9 2 ±4.6 3 ±4 4 ±3 5 0 Next Previous Main Menu A Close Look at Graphing End
PowerPoint Presentation : Graphing x 2 + y 2 = 25 x y Next Previous Main Menu A Close Look at Graphing End
PowerPoint Presentation : x 2 + y 2 + 6x = 16 x = 1 1 2 + y 2 +6(1) = 16 1 + y 2 +6 = 16 y 2 = 9 There are 2 points to graph: (1,3) (1,-3) y = ± 3 Next Previous Main Menu A Close Look at Graphing End
PowerPoint Presentation : Continue to solve in this manner, generating a table of values. -8 0 -7 ±3 -6 ±4 -5 ±4.6 -4 ±4.9 -3 ±5 -2 ±4.9 -1 ±4.6 0 ±4 1 ±3 2 0 -8 0 -7 ±3 -6 ±4 -5 ±4.6 -4 ±4.9 -3 ±5 -2 ±4.9 -1 ±4.6 0 ±4 1 ±3 2 0 Next Previous Main Menu A Close Look at Graphing End
PowerPoint Presentation : -8 0 -7 ±3 -6 ±4 -5 ±4.6 -4 ±4.9 -3 ±5 -2 ±4.9 -1 ±4.6 0 ±4 1 ±3 2 0 Graphing x 2 + y 2 + 6x = 16 Next Previous Main Menu A Close Look at Graphing End
PowerPoint Presentation : x 2 + y 2 - 4y = 21 x = 3 3 2 + y 2 - 4y = 21 9 + y 2 - 4y = 21 y 2 - 4y -12 = 0 There are 2 points to graph: (3,6) (3,-2) y = 6 and y = -2 (y - 6)(y + 2) = 0 y - 6 = 0 and y + 2 = 0 Next Previous Main Menu A Close Look at Graphing End
PowerPoint Presentation : Continue to solve in this manner, generating a table of values. 0 -3, 7 3 -2, 6 4 -1, 5 5 2 1 -2.9, 6.9 2 -2.6, 6.6 -5 2 -4 -1, 5 -3 -2, 6 -2 -2.6, 6.6 -1 -2.9, 6.9 0 -3, 7 3 -2, 6 4 -1, 5 5 2 1 -2.9, 6.9 2 -2.6, 6.6 -5 2 -4 -1, 5 -3 -2, 6 -2 -2.6, 6.6 -1 -2.9, 6.9 Next Previous Main Menu A Close Look at Graphing End
PowerPoint Presentation : Graphing x 2 + y 2 - 4y = 21 0 -3, 7 3 -2, 6 4 -1, 5 5 2 1 -2.9, 6.9 2 -2.6, 6.6 -5 2 -4 -1, 5 -3 -2, 6 -2 -2.6, 6.6 -1 -2.9, 6.9 Next Previous Main Menu A Close Look at Graphing End
PowerPoint Presentation : x 2 + y 2 + 6x - 4y = 12 x = 1 1 2 + y 2 + 6(1) - 4y = 12 1 + y 2 + 6 - 4y = 12 y 2 - 4y - 5 = 0 There are 2 points to graph: (1,5) (1,-1) y = 5 and y = -1 (y - 5)(y + 1) = 0 y - 5 = 0 and y + 1 = 0 Next Previous Main Menu A Close Look at Graphing End
PowerPoint Presentation : Continue to solve in this manner, generating a table of values. -8 2 -7 -1, 5 -6 -2, 6 -5 -2.6, 6.6 -3 -3, 7 -2 -2.9, 6.9 -1 -2.6, 6.6 -4 -2.9, 6.9 0 -2, 6 1 -1, 5 2 2 -8 2 -7 -1, 5 -6 -2, 6 -5 -2.6, 6.6 -3 -3, 7 -2 -2.9, 6.9 -1 -2.6, 6.6 -4 -2.9, 6.9 0 -2, 6 1 -1, 5 2 2 Next Previous Main Menu A Close Look at Graphing End
PowerPoint Presentation : Graphing x 2 + y 2 + 6x - 4y = 21 -8 2 -7 -1, 5 -6 -2, 6 -5 -2.6, 6.6 -3 -3, 7 -2 -2.9, 6.9 -1 -2.6, 6.6 -4 -2.9, 6.9 0 -2, 6 1 -1, 5 2 2 Next Previous Main Menu A Close Look at Graphing End
PowerPoint Presentation : x 2 + y 2 = 25 x 2 + y 2 + 4x = 16 x 2 + y 2 - 6y = 21 x 2 + y 2 + 4x - 6y = 12 What conclusions can you draw about the shape and location of the graphs? Next Previous Main Menu A Close Look at Graphing End
PowerPoint Presentation : What conclusions can you draw about the shape and location of the graphs? All of the graphs are circles. x 2 + y 2 = 25 As you add an x-term, the graph moves left. x 2 + y 2 + 4x = 16 As you subtract a y-term, the graph moves up. x 2 + y 2 - 6y = 21 You can move both left and up. x 2 + y 2 + 4x - 6y = 12 Next Previous Main Menu A Close Look at Graphing End
PowerPoint Presentation : A circle is the set of all points in a plane equidistant from a fixed point called the center . Center Circle Circle Center GEOMETRICAL DEFINITION OF A CIRCLE Next Previous Main Menu End
PowerPoint Presentation : A radius is the segment whose endpoints are the center of the circle, and any point on the circle. radius radius Next Previous Main Menu End
PowerPoint Presentation : A unit circle is a circle with a radius of 1 whose center is at the origin . It is the circle on which all other circles are based. A unit circle is a circle with a radius of 1 whose center is at the origin . r = 1 A unit circle is a circle with a radius of 1 whose center is at the origin . Center: (0, 0) A unit circle is a circle with a radius of 1 whose center is at the origin . Next Previous Main Menu End
PowerPoint Presentation : The equation of a circle is derived from its radius . Next Previous Main Menu End Last Viewed
PowerPoint Presentation : Use the distance formula to find an equation for x and y. This equation is also the equation for the circle . Next Previous Main Menu Deriving the Equation of a Circle End
PowerPoint Presentation : THE DISTANCE FORMULA (x 2 , y 2 ) (x 1 , y 1 ) Next Previous Main Menu Deriving the Equation of a Circle End
PowerPoint Presentation : Deriving the Equation of a Circle (x, y) (0, 0) Let r for radius length replace D for distance. r 2 = x 2 + y 2 r 2 = x 2 + y 2 Is the equation for a circle with its center at the origin and a radius of length r. Next Previous Main Menu End
PowerPoint Presentation : The unit circle therefore has the equation: x 2 + y 2 = 1 (x, y) (0, 0) r = 1 Next Previous Main Menu End Deriving the Equation of a Circle
PowerPoint Presentation : x 2 + y 2 = 4 If r = 2 , then (x, y) (0, 0) r = 2 Next Previous Main Menu Deriving the Equation of a Circle End
PowerPoint Presentation : In order for a satellite to remain in a circular orbit above the Earth, the satellite must be 35,000 km above the Earth. Write an equation for the orbit of the satellite. Use the center of the Earth as the origin and 6400 km for the radius of the earth. (x - 0) 2 + (y - 0) 2 = (35000 + 6400) 2 x 2 + y 2 = 1,713,960,000 Next Previous Main Menu Deriving the Equation of a Circle End
PowerPoint Presentation : What will happen to the equation if the center is not at the origin? (1,2) r = 3 (x,y) Next Previous Main Menu End Last Viewed
PowerPoint Presentation : No matter where the circle is located, or where the center is, the equation is the same. (h,k) (x,y) r (h,k) (x,y) r (h,k) (x,y) r (h,k) (x,y) r (h,k) (x,y) r Next Previous Main Menu Moving the Center End
PowerPoint Presentation : (x,y) Assume (x, y) are the coordinates of a point on a circle. (h,k) The center of the circle is (h, k) , r and the radius is r . Then the equation of a circle is: ( x - h ) 2 + ( y - k ) 2 = r 2 . Next Previous Main Menu Moving the Center End
PowerPoint Presentation : ( x - h ) 2 + ( y - k ) 2 = r 2 ( x - 0 ) 2 + ( y - 3 ) 2 = 7 2 (x) 2 + (y - 3) 2 = 49 Write the equation of a circle with a center at (0, 3) and a radius of 7 . Next Previous Main Menu Moving the Center End
PowerPoint Presentation : Find the equation whose diameter has endpoints of (-5, 2) and (3, 6). First find the midpoint of the diameter using the midpoint formula. MIDPOINT This will be the center . Next Previous Main Menu Moving the Center End
PowerPoint Presentation : Find the equation whose diameter has endpoints of (-5, 2) and (3, 6). Then find the length distance between the midpoint and one of the endpoints. DISTANCE FORMULA This will be the radius . Next Previous Main Menu Moving the Center End
PowerPoint Presentation : Find the equation whose diameter has endpoints of (-5, 2) and (3, 6). Therefore the center is (-1, 4) The radius is (x - -1) 2 + (y - 4) 2 = (x + 1) 2 + (y - 4) 2 = 20 Next Previous Main Menu Moving the Center Skip Tangents End
PowerPoint Presentation : A line in the plane of a circle can intersect the circle in 1 or 2 points. A line that intersects the circle in exactly one point is said to be tangent to the circle. The line and the circle are considered tangent to each other at this point of intersection. Write an equation for a circle with center (-4, -3) that is tangent to the x-axis. A diagram will help. (-4, 0) (-4, -3) 3 A radius is always perpendicular to the tangent line. (x + 4) 2 + (y + 3) 2 = 9 Next Previous Main Menu Moving the Center End
PowerPoint Presentation : The standard form equation for all conic sections is: Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 where A, B, C, D, E and F represent constants and where the equal sign could be replaced by an inequality sign. How do you put a standard form equation into graphing form? The transformation is accomplished through completing the square . Next Previous Main Menu End Last Viewed
PowerPoint Presentation : Graph the relation x 2 + y 2 - 10x + 4y + 13 = 0 . 13 1. Move the F term to the other side. x 2 + y 2 - 10x + 4y = - 13 3. Complete the square for the x-terms and y-terms . x 2 - 10x + y 2 + 4y = -13 x 2 - 10x + 25 + y 2 + 4y + 4 = -13 + 25 + 4 (x - 5) 2 + (y + 2) 2 = 16 x 2 + y 2 - 10x + 4y + 13 = 0 x 2 + y 2 - 10x + 4y = -13 2. Group the x-terms and y-terms together x 2 - 10x + y 2 + 4y = -13 Graph the relation Next Previous Main Menu Completing the Square End
PowerPoint Presentation : (x - 5) 2 + (y + 2) 2 = 16 center: ( 5 , -2 ) radius = 4 (x - 5) 2 + (y + 2) 2 = 4 2 4 (5, -2) Next Previous Main Menu Completing the Square End
PowerPoint Presentation : What if the relation is an inequality? x 2 + y 2 - 10x + 4y + 13 < 0 Do the same steps to transform it to graphing form. (x - 5) 2 + (y + 2) 2 < 4 2 This means the values are inside the circle. The values are less than the radius . Next Previous Main Menu Completing the Square End
PowerPoint Presentation : Write x 2 + y 2 + 6x - 2y - 54 = 0 in graphing form. Then describe the transformation that can be applied to the graph of x 2 + y 2 = 64 to obtain the graph of the given equation. x 2 + y 2 + 6x - 2y = 54 x 2 + 6x + y 2 - 2y = 54 ( 6 / 2 ) = 3 ( -2 / 2 ) = -1 (3) 2 = 9 (-1) 2 = 1 x 2 + 6x + 9 + y 2 - 2y + 1 = 54 + 9 + 1 (x + 3) 2 + (y - 1) 2 = 64 (x + 3) 2 + (y - 1) 2 = 8 2 center: (-3, 1) radius = 8 Next Previous Main Menu Completing the Square End
PowerPoint Presentation : Write x 2 + y 2 + 6x - 2y - 54 = 0 in graphing form. Then describe the transformation that can be applied to the graph of x 2 + y 2 = 64 to obtain the graph of the given equation. x 2 + y 2 = 64 is translated 3 units left and one unit up to become (x + 3) 2 + (y - 1) 2 = 64 . Next Previous Main Menu Completing the Square End
PowerPoint Presentation : The graph of a quadratic relation will be a circle if the coefficients of the x 2 term and y 2 term are equal (and the xy term is zero). Previous Main Menu End