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Transcript
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
x2  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