Circles Revision
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Transformations
Intercepts
Using the discriminant
Chords
From the circle: x2 + y2 = 1
to the circle: (x-1)2 + (y+3)2 = 9
What transformations have occurred?
x2 + y2 = 1
Centre (0,0)
Radius 1
x2 + y2 = 32
(x-1)2 + (y+3)2 = 9
Centre (0,0)
Radius 3
ENLARGED BY
SCALE FACTOR 3
(1,-3)
Centre (1,-3)
Radius 3
TRANSLATED BY
1
-3
[]
Where do they intersect?
For the circle: (x-1)2 + (y-3)2 = 9
.. and the line y = x +10
Where do they cross?
Solve simultaneously to find intersect …
2 – 2x +1) +
2
2
(x
(x-1) + (y-3) = 9
2 + 14x + 49) = 9
(x
y = x +10
Substitute y:
2x2 + 12x + 41 = 0
(x-1)2 + (x +10 -3)2 = 9
Solve equation to
find intersect
(x-1)2 + (x +7)2 = 9
Circle Intersect
Does 2x2 + 12x+41=0 have real roots
Use discriminent “b2-4ac”
To find out about roots
a = [coefficient of x2] = 2
b = [coefficient of x] = 12
c= [constant]
= 41
b2 - 4ac
= 122 – (4 x 2 x 41)
= 144 – 328
= -184
“b2 – 4ac < 0”
No roots
(solutions)
Circle Intersect
(x-1)2 + (y-3)2 = 9
y = x +10
2x2 + 12x+41=0 has no real roots
-> no solutions; so lines do not cross
What the
discriminent
tells us ……
“b2-4ac > 0”
- two solutions
- crosses
“b2-4ac = 0”
- one solution
- tangent
“b2-4ac < 0”
- no solutions
- misses
The really important stuff you need to know
about chords - but were afraid to ask
A chord joins any 2 points on a
circle and creates a segment
centre
The perpendicular bisector
of a chord passes through the
centre of the circle
… conversely a radius of the
circle passing perpendicular to
the chord will bisect it
Using these facts we can solve circle problems
(11,14)
Given these 2 chords …
find the centre of the circle
The perpendicular bisector
of a chord passes through the
centre of the circle
(5,10)
(4,7)
(8,3)
If you find the equation of
the two perpendicular
bisectors, where they cross
is the centre
Given these 2 chords …
find the centre of the circle
B (11,14)
M
Gradient of AB is : 14 - 10
11 - 5
= 4/6 = 2/3
A
(5,10)
(4,7)
Midpoint (M) of AB is …
(5 + 11 , 10 + 14) = (8, 12)
2
2
C
R
Gradient MC x 2/3 = -1
Gradient MC = -3/2
Equations of form y-y1=m(x-x1)
S (8,3)
Line goes through (x1,y1) with gradient m
y - 12 = -3/2 (x - 8)
Equation of perpendicular y - 12 = -3/2 x + 12
bisector of AB is: y = -3/2 x + 24
Given these 2 chords …
find the centre of the circle
B (11,14)
Midpoint (N) of RS is …
(4 + 8 , 7 + 3 ) = (6, 5)
2
2
Gradient of RS is :
A
= 4/-4 = -1
(5,10)
C
(4,7)
R
N
7-3
4-8
Gradient NC x -1 = -1
Gradient NC = 1
Equations of form y-y1=m(x-x1)
S (8,3)
Line goes through (x1,y1) with gradient m
y - 5 = 1 (x - 6)
Equation of perpendicular y - 5 = x - 6
bisector of RS is: y = x - 1
Finding the centre ….
y = -3/2 x + 24
(11,14)
If you find the equation of
the two perpendicular
bisectors, where they cross
is the centre
y=x-1
-
y = -3/2 x + 24
(5,10)
(4,7)
subtract
0 = x - -3/2x -1 - 24
C
5/2 x -25 = 0
5/2 x = 25
5x = 50
x = 10
y=x-1
y=x-1
(8,3)
y = 10 -1 = 9
Centre is at (10,9)