SPECIALIST MATHEMATICS UNIT 3: COORDINATE GEOMETRY - CIRCLES ‘Essential Specialist Mathematics’ text book (3rd edition) was used for the selection of questions below. TI Nspire CAS OS 3.2.1 Exercise 1F Question 3 f. Sketch the graph of the following: 3x 2 3 y 2 6 x 9 y 100 We may want to change from the general form to centre-radius form first: 3x 2 3 y 2 6 x 9 y 100 3 439 ( x 1)2 ( y )2 2 12 or 3 1317 ( x 1) 2 ( y ) 2 2 6 439 3 ( x 1)2 12 2 Use the list {1,-1} expression for the sign to draw and to define it as let say f. The calculator will find the y-intercepts by using substitution. However, it will not find x-intercepts by solving f 0 . Note that f is the relationship, not the function. You need to define a new expression here, f1, being the lower semi-circle and then solve f1 for zero. See the screens below. To draw it, make y the subject y Menu: 3: Graph Entry 3: Circle, there are two options. You can enter the equation in general form as 2: or in centre-radius form as 1: Use the list {1,-1} expression for the sign to draw the circle when using option 1: above. Note: by default there is a minus sign in option 1 so you need to enter ( x (1))2 , which is a little bit tricky. You can determine the centre and radius by Menu 6: Analyze Graph 8: Analyze Conics 1: Centre and 7: Radius 1 Define the circle as f1(x). Find yintercepts. Define the lower semi-circle as f2(x). Solve for zero to find the x-intercepts. Mode in Exact if exact answers required. Question 7 Find the equation of the circle which passes through (3, 1), (8,2) and (2,6). Use the form x 2 y 2 Dx Ey F 0 to create 3 simultaneous equations by substituting the coordinates of the points: 10 3D E F 0 68 8D 2 E F 0 40 2 D 6 E F 0 2 Then use your calculator to solve and substitute the values back into D E D2 E 2 4F to obtain the equation of the circle x y 2 2 4 2 2 ( x 5)2 ( y 4)2 13 You may wish to draw check that the 3 given points actually lay on the circle. Interpret the answers given by the calculator. Question 9b. Find the coordinates of the points of intersection of the circle with equation x 2 y 2 25 and the line with equation y 2 x. Using the calculator: The solutions are 5 2 5 and 5, 2 5 . You can enter equation of the circle and then change graph entry to 1: Function to enter the straight line. Both will appear in the same screen. Then you can find the points of intersection (numerical values only) in the screen. 3