Transcript Higher Outcome 4 - Official Mathematics Revision Website

Equation of a Circle
Centre at the Origin
By Pythagoras Theorem
x2  y2  r 2
y
y
O
P(x,y)
r
x
x
General Equation of a Circle
y
P(x,y)
y
b
C(a,b)
r
Circle centre C(a , b)
Radius r
y-b
x-a
O
a
x
x
By Pythagoras Theorem
( x  a)2  ( y  b) 2  r 2
When you are asked to determine the equation of
a circle, this is the equation you should use!
The Circle examples
(x – 2)2 + (y – 5)2 = 49
centre (2,5) radius = 7
(x + 5)2 + (y – 1)2 = 13
centre (-5,1) radius = 13
(x – 3)2 + y2 = 20
centre (3,0)
radius = 20
= 25
Centre (2,– 3) & radius = 10
Equation is (x – 2 )2 + (y + 3)2 = 100
Centre (0,6) & radius = 23
Equation is x2 + (y – 6 )2 = 12
P(5 , 2)
3
C
4
(2, – 2)
Find the equation of the circle
that has PQ as diameter
C, the centre is the mid point of PQ
Q
(– 1, – 6)
C (2,– 2)
CP2, Radius, r2 = 32 + 42
Using
= 25
(x – a)2 + (y – b)2 = r2
Equation is (x – 2)2 + (y + 2)2 = 25
Two circles are concentric, same centre
The larger has equation
(x + 3)2 + (y – 5)2 = 12
The radius of the smaller is
half that of the larger.
Find its equation.
From
(x – a)2 + (y – b)2 = r2
Centres is (– 3, 5)
Larger radius = 12
 4 3
Smaller radius = 3
so
= 2 3
r2 = 3
Required equation is (x + 3)2 + (y – 5 )2 = 3
Inside / Outside or On Circumference
When a circle has equation
(x – a)2 + (y – b)2 = r2
If (x,y) lies on the circumference then
– a)2 + (y – b)2 = r2
If (x,y) lies inside the circumference then
(x – a)2 + (y – b)2 < r2
If (x,y) lies outside the circumference then
(x – a)2 + (y – b)2 > r2
(x
Inside / Outside or On Circumference
Consider the circle (x + 1)2 + (y – 4)2 = 100
Determine where the following points lie;
K(–7,12) , L(10,5) , M(4,9)
Centre C(–1,4) , radius, r = 10  r2 = 100
CK2 = 62 + 82 = 100
CL2 = 112 + 12 = 122
CM2 = 52 + 52
= 50
 K on circumference
 L outside
 M inside
Another version of Circle Equation
(x – a)2 + (y – b)2 = r2
is a circle : centre (a , b) radius, r
x2 + y2 + 2gx + 2fy + c = 0
is a circle : centre (– g , – f) radius, r = √g2 + f2 - c
Equation x2 + y2 + 2gx + 2fy + c = 0
Write the equation (x – 5)2 + (y + 3)2 = 49 without
brackets.
(a + b)2
(x – 5)2 + (y + 3)2 = 49
= a2 + 2ab + b2
x2 – 10x + 25 + y2 + 6y + 9 – 49 = 0
x2 + y2 – 10x + 6y – 15 = 0
This takes the form given above where
2g = – 10 , 2f = 6 and c = – 15
Show that the equation
x2 + y2 - 6x + 2y - 71 = 0
represents a circle and find the centre and radius.
Circle exists if radius exists
r 
g  f c
2
2
x2 + y2 – 6x + 2y – 71 = 0
x2 + y2 + 2gx + 2fy + c = 0
So g = – 3, f = 1, c = – 71
r 
32  12  71
r 
81
r 9
So equation represents a circle radius 9
Centre is (– g, – f)
x2 + y2 - 6x + 2y - 71 = 0
Show that the equation
represents a circle and find the centre and radius.
Circle exists if radius exists and the coefficients of
x2 and y2 are equal
r 
g  f c
2
2
x2 + y2 – 6x + 2y – 71 = 0
x2 + y2 + 2gx + 2fy + c = 0
So g = – 3, f = 1, c = – 71
r 
32  12  71
r 
81
r 9
So equation represents a circle radius 9
Centre is (– g, – f)
 Centre is (3, – 1)
We now have 2 ways on finding the centre and
radius of a circle depending on the form we have.
x2 + y2 – 10x + 6y – 15 = 0
x2 + y2 + 2gx + 2fy + c = 0
2g = – 10
g=–5
centre = (-g,-f)
= (5,-3)
2f = 6
f=3
c = – 15
radius = (g2 + f2 – c)
= (25 + 9 – (-15))
= 49
= 7
x2 + y2 – 6x + 2y – 71 = 0
x2 + y2 + 2gx + 2fy + c = 0
2g = -6
2f = 2
g = -3
f=1
centre = (-g,-f)
= (3,-1)
c = -71
radius = (g2 + f2 – c)
= (9 + 1 – (-71))
= 81
= 9
Find the centre & radius of
x2 + y2 – 10x + 4y – 5 = 0
x2 + y2 – 10x + 4y – 5 = 0
x2 + y2 + 2gx + 2fy + c = 0
2g = -10
g = -5
centre = (-g,-f)
= (5,-2)
2f = 4
c = -5
f=2
radius = (g2 + f2 – c)
= (25 + 4 – (-5))
= 34
The circle x2 + y2 – 10x – 8y + 7 = 0
cuts the y- axis at A & B. Find the length of AB.
y
A
At A & B x = 0
so
the equation becomes
y2 – 8y + 7 = 0
(y – 1)(y – 7) = 0
B
x
y = 1 or y = 7
A is (0,7) & B is (0,1)
So AB = 6 units
Frosty the Snowman’s lower body section can be
represented by the equation x2 + y2 – 6x + 2y – 26 = 0
His middle section is the same size as the lower
but his head is only 1/3 the size of the other two
sections. Find the equation of his head !
x2 + y2 – 6x + 2y – 26 = 0
2g = -6
g = -3
2f = 2
f=1
centre = (-g,-f)
= (3,-1)
c = -26
radius = (g2 + f2 – c)
= (9 + 1 + 26)
= 36
= 6
(3,19)
2
6
(3,11)
radius of head = 1/3 of 6 = 2
Using (x – a)2 + (y – b)2 = r2
6
6
(3,-1)
Equation is
(x – 3)2 + (y – 19)2 = 4
By considering centres and radii prove that the
following two circles touch each other.
Circle 1
x2 + y2 + 4x – 2y – 5 = 0
Circle 2
x2 + y2 – 20x + 6y + 19 = 0
Circle 1
2g = 4 so g = 2
2f = -2 so f = -1
c = -5
centre = (-g, -f) = (-2,1)
radius =
(g2
+
f2
– c)
= (4 + 1 + 5)
= 10
Circle 2
2g = -20 so g = -10
2f = 6 so f = 3 c = 19
centre = (-g, -f) = (10,-3)
radius = (g2 + f2 – c)
= (100 + 9 – 19)
= 90
= 310
(-2,1)
√10
(10,-3)
3√10
If d is the distance
between the centres
then by Pythagoras
d2 = 122 + 42
= 144 + 16
radius 1 + radius 2
= 10 + 310
= 160
= 16 × 10
d = 410
= 410
= distance between centres
It now follows that the circles must touch !
Intersection of Lines & Circles
There are 3 possible scenarios
2 points of
contact
(b2- 4ac > 0)
1 point of
contact
(b2- 4ac = 0)
line is tangent
0 points of
contact
(b2- 4ac < 0)
To determine where the line and circle meet we solve
simultaneous equations and the discriminant tells us
how many solutions we have.
Equations of Tangents
At the point of contact a tangent and radius /diameter are
perpendicular.
radius
Tangent
This means we make use of
m1m2 = -1.
Prove that the point (-4,4) lies on the circle
x2 + y2 – 12y + 16 = 0
Find the equation of the tangent here.
At (-4,4)
x2 + y2 – 12y + 16 = 16 + 16 – 48 + 16 = 0
So (-4,4) must lie on the circle.
x2 + y2 – 12y + 16 = 0
2g = 0 so g = 0
2f = -12 so f = -6
Centre is (-g,-f) = (0,6)
(0,6)
(-4,4)
Prove that the point (-4,4) lies on the circle
x2 + y2 – 12y + 16 = 0
Find the equation of the tangent here.
mradius = 2/4 = 1/2
m1× m2= – 1
(0,6)
mtangent = – 2
y – b = m(x – a)
4
-2
-4
y – 4 = -2(x + 4)
y – 4 = -2x – 8
2x + y + 4 = 0
(-4,4)
Find where the line y = 2x + 1 meets the circle
(x – 4)2 + (y + 1)2 = 20 and comment on the answer
Replace y by 2x + 1 in the circle equation
(x – 4)2 + (y + 1)2 = 20
becomes
(x – 4)2 + (2x + 1 + 1)2 = 20
(x – 4)2 + (2x + 2)2 = 20
x2 – 8x + 16 + 4x2 + 8x + 4 = 20
5x2 = 0
x = 0 one solution  tangent
Using y = 2x + 1, if x = 0 then y = 1
Point of contact is (0,1)
Find where the line y = 2x + 6 meets the circle
x2 + y2 + 10x – 2y + 1 = 0
Replace y by 2x + 6 in the circle equation
x2 + (2x + 6)2 + 10x – 2(2x + 6) + 1 = 0
x2 + 4x2 + 24x + 36 + 10x – 4x – 12 + 1 = 0
5x2 + 30x + 25 = 0
( 5 )
x2 + 6x + 5 = 0
(x + 5)(x + 1) = 0
 x = -5 or x = -1
Using y = 2x + 6
x = -5  y = -4
Points of contact
x = -1  y = 4
(-5,-4) and (-1,4)
Tangency
Prove that the line 2x + y = 19 is a tangent to the circle
x2 + y2 - 6x + 4y - 32 = 0 , and also find the point of contact.
2x + y = 19
so
y = 19 – 2x
Replace y by (19 – 2x) in the circle equation.
x2 + (19 – 2x)2 – 6x + 4(19 – 2x) - 32 = 0
x2 + 361 – 76x + 4x2 - 6x + 76 – 8x – 32 = 0
5x2 – 90x + 405 = 0
x2 – 18x + 81 = 0
(x – 9)(x – 9) = 0
x = 9 twice
 tangent
( 5)
Using
y = 19 – 2x
x=9  y=1
Point of contact
is (9,1)
Using Discriminants
At the line x2 – 18x + 81 = 0 we can also show there is
only one solution by showing that the discriminant is zero.
For x2 – 18x + 81 = 0 , a =1, b = -18 and c = 9
So
b2 – 4ac = (-18)2 – 4 × 1 × 81 = 364 - 364
=0
Since discriminant = 0 the equation has only one root
so there is only one point of contact so line is a tangent.
The next example uses discriminants in a slightly
different way.
Find the equations of the tangents to the circle
x2 + y2 – 4y – 6 = 0 from the point (0,-8).
Each tangent takes the form y = mx – 8
Replace y by (mx – 8) in the circle equation
x2 + (mx – 8)2 – 4(mx – 8) – 6 = 0
x2 + m2x2 – 16mx + 64 – 4mx + 32 – 6 = 0
(m2+ 1)x2 – 20mx + 90 = 0
Tangency  b2 – 4ac =0 see next slide
Find the equations of the tangents to the circle
x2 + y2 – 4y – 6 = 0 from the point (0,-8).
(m2+
1)x2
– 20mx + 90 = 0
Tangency 
b2
– 4ac =0
a = (m2 + 1)
b = -20m
c = 90
(-20m)2 – 4 × (m2 + 1) × 90 = 0
400m2 – 360m2 – 360 = 0
40m2 – 360 = 0
m2 = 9
m= ± 3
Equations of tangents
y = -3x – 8
y = 3x - 8
Quadratic Theory
Discriminant
Graph sketching
r  g2  f 2 c
Move the circle from the origin
a units to the right
b units upwards
b2 - 4ac  0
Used for
intersection problems
between circles and lines
NO intersection
b2 - 4ac  0
x2 y2 2g x2f  yc0
Centre
(-g,-f)
Centre
(a,b)
Factorisation
The
Circle
(x a)2 (y b)2  r2
Special case
x2  y2  r2
b2 - 4ac  0
line is a tangent
Pythagoras Theorem
Rotated
through 360 deg.
Centre
(0,0)
Mind Map
2 pts of intersection
For Higher Maths Topic : The Circle
Created by Mr. Lafferty
Two circles touch
externally if
the distance
Straight line Theory
Two circles touch
internally if
the distance
C1C2  (r2  r1)
C1C2  (r1  r2 )
Distance formula
2
2
CC

(
y

y
)

(
x

x
)
1 2
2
1
2
1
Perpendicular
equation
m1  m2  1
Find the equation of the circle with
centre(–3 , 4) and passing through the
origin.
Find radius (distance formula):
r 5
You know the centre:
Write down equation:
(3, 4)
( x  3)  ( y  4)  25
2
2
Explain why this equation does not represent a circle.
x2  y 2  2 x  3 y  5  0
Consider the 2 conditions
1. Coefficients of x2 and y2 must be the same.
2. Radius must be > 0
Calculate g and f:
Evaluate g  f  c
2
Deduction:
2
g  1,
f 
 
(1)  
2
i.e. g 2  f 2  c  0
3
2
2
5
g 2  f 2  c  0 so
3
2

1
4
1  2 5  0
g 2  f 2  c not real
Equation does not represent a circle
Find the equation of the circle which has P(–2, –1) and Q(4, 5)
as the end points of a diameter.
Q(4, 5)
Make a sketch
C
(1 , 2)
P(-2, -1)
Calculate mid-point for centre:
Calculate radius CQ:
(1, 2)
r2 = 32 + 32 = 18
Using (x – a)2 + (y – b)2 = r2
Write down equation:
 x  1
2
  y  2   18
2
Find the equation of the tangent at the point (3, 4) on the circle
x2  y 2  2x  4 y 15  0
P(3, 4)
x2 + y2 + 2gx + 2fy + c = 0
Find centre of circle: (-g , -f) (1, 2)
Make a sketch
mOP (radius to tangent)
Gradient of tangent:
O(-1, 2)
1
m
2
m  2
y – b = m(x – a)
Equation of tangent:
y – 4 = -2(x – 3)
 y = -2x + 10
O, A and B are the centres of the three circles shown in the
diagram. The two outer circles are congruent, each touches the
smallest circle. Circle centre A has equation (x – 12)2 + (y + 5)2 = 25
The three centres lie on a parabola whose axis of symmetry is
shown the by broken line through A.
a) i) State coordinates of A and find length of line OA.
ii) Hence find the equation of the circle with centre B.
b) The equation of the parabola can be written in the form
y = px(x + q)
find p and q.
(x – 12)2 + (y + 5)2 = 25
A(12 , -5)
OA2 = 122 + (-5)2
OA2 = 169
OA = 13
OB = 13
Radius small circle = 5
Radius circle B = 8
Centre circle B = (24 , 0),
by symmetry
Equation circle B is
(x – 24)2 + (y – 0)2 = 82
Equation circle B is
(x – 24)2 + y2 = 64
Equation of parabola is
y = px(x – 24), q = -24
Parabola passes through
A(12 , -5)
-5 = p × 12(12 – 24)
-5 = p × -144
p = 5/144
Circle P has equation x2 + y2 – 8x – 10y + 9 = 0.
centre (–2, –1) and radius 22.
Circle Q has
a) i) Show that the radius of circle P is 42
ii) Hence show that circles P and Q touch.
b) Find the equation of the tangent to circle Q at the point (–4, 1)
c) The tangent in (b) intersects circle P in two points.
Find the x-coordinates of the points of intersection, expressing
your answers in the form a ± b√3
x2 + y2 + 2gx + 2fy + c = 0
r  g2  f 2  c
x2 + y2 – 8x – 10y + 9 = 0
r  42  52  9
r  32  (16  2)
r4 2
If circles touch the sum of the
radii will equal the distance
between the centres
(-2 , -1) to (4 , 5)  d2 = 62 + 62
d2 = 72
d = 6√2
d2
P
(4 , 5)
(-4 , 1)
2√2
= 36 × 2
Q
So circles touch
(-2 , -1)
Tangent is perpendicular to radius
m radius = 2/-2 = -1
 m tangent = 1
y – b = m(x – a)
-4
1 1
 y – 1 = 1(x + 4)
y=x+5
4√2
Solve equations to find
points of intersection
x2 + y2 – 8x – 10y + 9 = 0
Replace y in circle
equation by (x + 5)
y=x+5
(-4 , 1)
2√2
P
(4 , 5)
4√2
Q
(-2 , -1)
x2 + (x + 5)2 – 8x – 10(x + 5) + 9 = 0
x2 + x2 + 10x + 25 – 8x – 10x – 50 + 9 = 0
2x2 – 8x – 16 = 0
(÷ by 2)
x2 – 4x – 8 = 0  a = 1, b = -4 and c = -8
x = 2 ± 2√3
 b  b 2  4ac
x
2a
4  16  32
x
2
4  48
x
2
4  16  3
x
2
44 3
x
2
For what range of values of c does the equation
x2 + y2 – 6x + 4y + c = 0 represent a circle ?
From x2 + y2 + 2gx + 2fy + c = 0
g = 3, f = -2 and c = c
Circle exists if radius exists and the coefficients of
x2 and y2 are equal
r2 = g2 + f2 - c
r2 = 32 + (-2)2 - c
r2 = 13 – c
For circle to exist 13 – c > 0
 c < 13
When newspapers were printed by lithograph, the newsprint had
to run over three rollers, illustrated in the diagram by 3 circles.
The centres A, B and C of the three circles are collinear.
The equations of the circumferences of the outer circles are
(x + 12)2 + (y + 15)2 = 25 and (x – 24)2 + (y – 12)2 = 100
Find the equation of the central circle.
A(-12 , -15)
10
r=5
C(24 , 12)
C(24 , 12)
r = 10
5
A(-12 , -15)
AC2 = 362 + 272
10
C(24 , 12)
45
Diameter of B is
45 – 15 = 30
15
15
5
AC2 = 2025
AC = 45
radius = 15
A(-12 , -15)
B divides AC in the ratio 20 : 25 = 4 : 5
So B is 4/9 along the line from A to C
x – coordinate: -12 to 24 = 36
4/
y – coordinate: -15 to 12 = 27
4/
B( 4 , -3)
9
9
of 36 = 16 xB = -12 + 16 = 4
of 27 = 12 yB = -15 + 12 = -3
Equation (x – 4)2 + (y + 3)2 = 152
The point P(2, –3) lies on the
circle with centre C as shown.
The gradient of CP is –2.
What is the equation of the
tangent at P?
y – b = m(x – a)
Tangent and radius are
perpendicular.
m1m2 = – 1
mtangent = 1/2
–3
1/
2
2
y + 3 = 1/2(x – 2)
Circle C1 has equation (x + 1)2 + (y – 1)2 = 121.
A circle C2 with equation x2 + y2 – 4x + 6y + p = 0 is drawn
inside C1.
The circles have no points of contact.
What is the range of values of p?
C1: centre (–1 , 1), r = 11
C2: centre (2 , –3), r = √(4 + 9 – p) = √(13 – p)
(-1 , 12)
(-11 , 1)
(-1 , 1) (10 , 1)
• (2 , -3)
(-1 , -10)