UNIT 6 MODELING GEOMETRY Lesson 1: Deriving Equations Instruction

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1 Prerequisite Skills This lesson requires the use of the following skills: appling the Pthagorean Theorem representing horizontal and vertical distances in a coordinate plane simplifing square roots writing equivalent forms of epressions involving squares and square roots writing equivalent forms of epressions involving perfect square trinomials and other polnomials completing the square to form perfect square trinomials Introduction The graph of an equation in and is the set of all points (, ) in a coordinate plane that satisf the equation. Some equations have graphs with precise geometric descriptions. For eample, the graph of the equation = + 3 is the line with a slope of, passing through the point (0, 3). This geometric description uses the familiar concepts of line, slope, and point. The equation = + 3 is an algebraic description of the line. In this lesson, we will investigate how to translate between geometric descriptions and algebraic descriptions of circles. We have alread learned how to use the Pthagorean Theorem to find missing dimensions of right triangles. Now, we will see how the Pthagorean Theorem leads us to the distance formula, which leads us to the standard form of the equation of a circle. Ke Concepts The standard form of the equation of a circle is based on the distance formula. The distance formula, in turn, is based on the Pthagorean Theorem. The Pthagorean Theorem states that in an right triangle, the square of the hpotenuse is equal to the sum of the squares of the legs. c a b a + b = c U6-8 CCGPS Analtic Geometr Teacher Resource

2 If r represents the distance between the origin and an point (, ), then + = r. (, ) (, ) r r (0, 0) (0, 0) can be positive, negative, or zero because it is a coordinate. can be positive, negative, or zero because it is a coordinate. r cannot be negative because it is a distance. A circle is the set of all points in a plane that are equidistant from a reference point in that plane, called the center. The set of points forms a -dimensional curve that measures 360º. The center of a circle is the point in the plane of the circle from which all points on the circle are equidistant. The center is in the interior of the circle. The radius of a circle is the distance from the center to a point on the circle. For a circle with center (0, 0) and radius r, an point (, ) is on that circle if and onl if + = r. (, ) (0, 0) r The distance formula is used to find the distance between an two points on a coordinate plane. The distance formula states that the distance d between A ( 1, 1 ) and B (, ) is ( ) + ( ) d= 1 1 The distance formula is based on the Pthagorean Theorem.. U6-9 CCGPS Analtic Geometr Teacher Resource

3 Look at this diagram of a right triangle with points A, B, and C. The distance d between points A and B is unknown. A ( 1, 1 ) 1 d C ( 1, ) 1 B (, ) The worked eample that follows shows how the distance formula is derived from the Pthagorean Theorem, using the points from the diagram to find d: AB = BC + AC Pthagorean Theorem d = d = Substitute values for sides AB, BC, and AC of the triangle. ( 1) + ( 1) Simplif. All squares are nonnegative. ( ) + ( ) d= 1 1 Take the square of each side of the equation to arrive at the distance formula. U6- CCGPS Analtic Geometr Teacher Resource

4 For a circle with center (h, k) and radius r, an point (, ) is on that circle if and onl if ( ) + ( ) = h k r. Squaring both sides of this equation ields the standard form of an equation of a circle with center (h, k) and radius r: ( h) + ( k) = r. (, ) r k (h, k) h (, k) If a circle has center (0, 0), then its equation is ( 0) + ( 0) = r, or + = r. If the center and radius of a circle are known, then either of the following two methods can be used to write an equation for the circle: Appl the Pthagorean Theorem to derive the equation. Or, substitute the center coordinates and radius directl into the standard form. The general form of an equation of a circle is A + B + C + D + E = 0, where A = B, A 0, and B 0. If an one of the following three sets of facts about a circle is known, then the other two can be determined: center (h, k) and radius r standard equation: ( h) + ( k) = r general equation: A + B + C + D + E = 0 U6-11 CCGPS Analtic Geometr Teacher Resource

5 The general form of the equation of a circle comes from epanding the standard form of the equation of the circle. The standard form of the equation of a circle comes from completing the square from the general form of the equation of a circle. b Ever perfect square trinomial has the form + b + because it is the square of a b b binomial: + b + = +. b Completing the square is the process of determining the value of and adding it to + b b to form the perfect square trinomial, + b +. Common Errors/Misconceptions confusing the radius with the square of the radius forgetting to square half the coefficient of when completing the square neglecting to square the denominator when squaring a fraction U6-1 CCGPS Analtic Geometr Teacher Resource

6 Guided Practice Eample 1 Derive the standard equation of the circle with center (0, 0) and radius Sketch the circle. 8 6 (, ) U6-13 CCGPS Analtic Geometr Teacher Resource

7 . Use the Pthagorean Theorem to derive the standard equation. In order to use the Pthagorean Theorem, there must be a right triangle. To create a right triangle, draw a line from point (, ) that is perpendicular to the horizontal line through the circle. Label the resulting sides of the triangle and. 8 6 (, ) Substitute the values for each side of the triangle into the formula for the Pthagorean Theorem, a + b + c. a + b + c + = 5 + = 5 Pthagorean Theorem Substitute values from the triangle. Simplif. The standard equation is + = 5. U6-1 CCGPS Analtic Geometr Teacher Resource

8 Eample Derive the standard equation of the circle with center (, 1) and radius. Then use a graphing calculator to graph our equation. 1. Sketch the circle. 8 6 (, ) (, 1) Use the Pthagorean Theorem to derive the standard equation. Create a right triangle. Draw lines from point (, ) and point (, 1) that meet at a common point and are perpendicular to each other. The length of the base of the triangle is equal to the absolute value of the difference of the -coordinates of the endpoints. The height of the triangle is equal to the absolute value of the difference of the -coordinates of the endpoints. (continued) U6-15 CCGPS Analtic Geometr Teacher Resource

9 8 6 (, ) 1 (, 1) (, 1) Substitute the resulting values for the sides of the triangle into the Pthagorean Theorem. a + b + c Pthagorean Theorem + 1 = Substitute values from the triangle. ( ) + ( 1) = All squares are nonnegative, so replace the absolute value smbols with parentheses. ( ) + ( 1) = 16 Simplif. The standard equation is ( ) + ( 1) = 16. U6-16 CCGPS Analtic Geometr Teacher Resource

10 3. Solve the standard equation for to obtain functions that can be graphed. ( ) + ( 1) = 16 Standard equation ( 1) = 16 ( ) Subtract ( ) from both sides. ( ) 1=± 16 If a = b, then a=± b. ( ) = 1± 16 Add 1 to both sides to solve for. ( ). Now graph the two functions, = and = On a TI-83/8: ( ) Step 1: Press [Y=]. Step : At Y 1, tpe in [1][+][ ][16][ ][(][X, T, θ, n][ ][][)][ ][)]. Step 3: At Y, tpe in [1][ ][ ][16][ ][(][X, T, θ, n][ ][][)][ ][)]. Step : Press [WINDOW] to change the viewing window. Step 5: At Xmin, enter [( )][9]. Step 6: At Xma, enter [9]. Step 7: At Xscl, enter [1]. Step 8: At Ymin, enter [( )][6]. Step 9: At Yma, enter [6]. Step : At Yscl, enter [1]. Step 11: Press [GRAPH]. On a TI-Nspire: Step 1: Press the [home] ke. Step : Arrow to the graphing icon and press [enter]. Step 3: At the blinking cursor at the bottom of the screen, tpe [1][+] [ ][16][ ][(][][ ][][)][ ][enter]. (continued) U6-17 CCGPS Analtic Geometr Teacher Resource

11 Step : Move the cursor to the bottom left of the screen and click on the double right-facing arrows. Step 5: At the blinking cursor, tpe [1][ ][ ][16][ ][(][][ ][][)][ ] [)][enter]. Step 6: Change the viewing window b pressing [menu], using the arrows to navigate down to number : Window/Zoom, and clicking the center button of the navigation pad. Step 7: Choose 1: Window settings b pressing the center button. Step 8: Enter in an appropriate XMin value, 9, b pressing [( )] and [9], then press [tab]. Step 9: Enter in an appropriate XMa value, [9], then press [tab]. Step : Leave the XScale set to Auto. Press [tab] twice to navigate to YMin and enter an appropriate YMin value, 6, b pressing [( )] and [6]. Step 11: Press [tab] to navigate to YMa. Enter [6]. Press [tab] twice to leave YScale set to auto and to navigate to OK. Step 1: Press [enter]. Step 13: Press [menu] and select : View and 5: Show Grid U6-18 CCGPS Analtic Geometr Teacher Resource

12 Eample 3 Write the standard equation and the general equation of the circle that has center ( 1, 3) and passes through ( 5, 5). 1. Sketch the circle. 8 ( 5, 5) r 6 ( 1, 3) U6-19 CCGPS Analtic Geometr Teacher Resource

13 . Use the distance formula to find the radius, r. ( ) + ( ) d= 1 1 Distance formula Substitute ( 1, 3) and ( 5, 5) for r = [( 5) ( 1) ] + ( 5 3) ( 1, 1 ) and (, ). r = Simplif. ( ) +( ) r = 16+ r = 0 r = r = 5 r = 5 5 Write 0 as a product with a perfect square factor. Appl the propert ab = a b. Simplif. 3. Substitute the center and radius directl into the standard form. ( h) + ( k) = r Standard form ( ) ( ) = ( ) ( + 1) + ( 3) = 0 The standard equation is ( + 1) + ( 3) = 0. Substitute values into the equation, using the center ( 1, 3), and the radius 5. Simplif to obtain the standard equation. U6-0 CCGPS Analtic Geometr Teacher Resource

14 . Square the binomials and rearrange terms to obtain the general form. ( + 1) + ( 3) = 0 Standard equation ( + 1)( + 1) + ( 3)( 3) = 0 Epand the factors = = = = 0 The general equation is = 0. Square the binomials to obtain trinomials. Combine the constant terms on the left side of the equation. Subtract 0 from both sides to get 0 on the right side. Rearrange terms in descending order to obtain the general equation. U6-1 CCGPS Analtic Geometr Teacher Resource

15 Eample Find the center and radius of the circle described b the equation = Rewrite the equation in standard form = = General form of the equation Subtract from both sides to get the constant term on one side = Group same-variable terms. Net, complete the square for both variables. Add the same values to both sides of the equation as shown: = = Simplif the equation shown above. ( ) + ( + 1) = 15 Write the perfect square trinomials as squares of binomials. The standard equation is ( ) + ( + 1) = 15.. Determine the center and radius. Write the standard equation ( ) + ( + 1) = 15 from step 1. ( ) + ( 1) = Rewrite to match the form ( 15) ( h) + ( k) = r. For the equation ( h) + ( k) = r, the center is (h, k) and the radius ( ) + ( ) is r, so for the equation 1 15, the center is (, 1) and the radius is 15. = ( ) U6- CCGPS Analtic Geometr Teacher Resource

16 Eample 5 Find the center and radius of the circle described b the equation = Rewrite the equation in standard form = = = = 9 General form of the equation Divide each term on both sides b to make the leading coefficient 1. Subtract 9 from both sides to get the constant term on one side. Combine like terms. Net, complete the square for both variables. Add the same values to both sides of the equation, as shown below: = = Simplif the equation from above ( 5 ) = 9 Write the perfect square trinomials as squares of binomials. 5 The standard equation is ( ) = 9. U6-3 CCGPS Analtic Geometr Teacher Resource

17 . Determine the center and radius ( ) 5 = ( ) = 3 5 Write the standard equation from step 1. Rewrite to match the form ( h) + ( k) = r. For the equation ( h) + ( k) = r, the center is (h, k) and the radius is r, so for the equation , 5 and the radius is 3. 3 ( ) =, the center is U6- CCGPS Analtic Geometr Teacher Resource