Transcript Book 6 Chapter 23 Locus - Wah Yan College, Kowloon
23
Locus
Case Study
23.1 Concept of Loci
23.2 Equations of Circles
23.3 Intersection of a Straight Line and a Circle
Chapter Summary
Case Study
Do you know how the Chang’e 1 Satellite orbited the Moon?
The satellite should move in a way that prevents itself from deviating from the path.
On 24 October 2007, China’s first lunar orbited ‘Chang’e 1 Satellite’ was launched from Xichuan.
The satellite orbited the Moon at an altitude of 200 km above the lunar surface.
It carried out a one-year lunar exploration mission.
P. 2
23.1 Concept of Loci
A. Description of Loci
If a point is moving under a specific condition, then its path is called the
locus
of the moving point. Consider a football rolling on a horizontal and smooth ground, the
locus
of the centre of the football is a straight line as shown in the figure below.
When a pendulum swings to and fro, the locus of its tip is an arc of a circle with the length of the pendulum as the radius. In mathematics, locus is the set of points that satisfy or are determined by some specific conditions. It can be a straight line, curve, polygon or circle.
P. 3
23.1 Concept of Loci
A. Description of Loci
Case 1 :
A point
R
moves in a plane such that it always maintains a fixed distance of 3 cm from a fixed line
L
.
Locus :
The locus of the point
R
is a pair of straight lines which are
(1) (2)
parallel to
L
; on one of the sides of
L
respectively;
(3)
equidistant from
L
with a distance of 3 cm.
Case 2 :
A point
Q
moves in a plane such that it always maintains an equal distance of 2 cm from two parallel lines
AB
and
CD
.
Locus :
The locus of the point
Q
is a straight line which is
(1) (2) (3)
parallel to both
AB
and
CD
; in the middle of
AB
and
CD
; equidistant from
AB
and
CD
with a distance of 2 cm.
P. 4
23.1 Concept of Loci
A. Description of Loci
Example 23.1T
Describe the locus of the moving point
P
under each of the following conditions. (a) (b)
P
is always 3 cm from the origin
O
.
AB
is a straight line and
PAB
PBA
.
Solution:
(a) A circle centred at
O
with radius 3 cm.
P
O
(b) The perpendicular bisector of
AB
.
P. 5
23.1 Concept of Loci
A. Description of Loci
Example 23.2T
In the figure,
l
1 point
P
and
l
2 are two parallel lines 6 cm apart. A moving is equidistant from
l
1 and
l
2 . Sketch the locus of
P
.
Solution:
The locus is a line which is parallel to
l
1 and and
l
2 .
l
2 , and is 3 cm from
l
1
P. 6
23.1 Concept of Loci
A. Description of Loci
Example 23.3T
In the figure,
AB
is a straight line. The locus of a point
P
is formed by the centres of all the circles with radius 1 cm and touch
AB
. Describe the locus of
P
.
Solution:
A pair of lines parallel to
AB
and are 1 cm away from it.
P. 7
23.1 Concept of Loci
A. Description of Loci
Example 23.4T
AB
is a line segment. A moving point and moves in a way such that
APB
P
passes through
A
130 ° . Describe and and
B
sketch the locus of
P
.
Solution:
The locus of
P
is a minor arc
APB
of a circle.
P. 8
23.1 Concept of Loci
B. Describe Loci in Algebraic Equations
Besides verbal description of the locus, we can also describe the locus of a point in an algebraic equation.
P. 9
23.1 Concept of Loci
B. Describe Loci in Algebraic Equations
Example 23.5T
A moving point
P
is always 2 units from Express the locus of
P
in algebraic form.
Q
1 2 , 1 .
Solution:
Let (
x
,
y
) be the coordinates of
P
. Since the distance between
P
and
Q
is 1 2 2 (
y
1 ) 2
x
2 2 1 (
y
1 ) 2 2 2
x
1 4
x
2
y
2
y
2 2
y
x
2
y
3 4 1 2 0 2 1 2 2 (
y
1 ) 2 , The locus of
P
4
x
2 + 4
y
2 – 4
x
is + 8
y
– 3 0.
P. 10
23.1 Concept of Loci
B. Describe Loci in Algebraic Equations
Example 23.6T
Consider a point
K
(5, 0) and a horizontal line
y
–2.
P
is a moving point such that its distance from the line is equal to its distance from the point
K
. Express the locus of
P
in the form
y
ax
2 +
bx
+
c
.
Solution:
Let (
x
,
y
) be the coordinates of
P
.
Since the distance from
P
to the line is equal to the distance from
P
to the point (
x
, 2) on the line, distance from
P
to (
x
, 2) distance from
P
to
K
(5, 0) (
x
x
) 2 [
y y
2 (
y
( 2 )] 2 2 ) 2 4
y
4 4
y
(
x
(
x
5 ) 2 5 ) 2 (
y
0 ) 2
y
2
x x
2 2 10
x
10
x
25 21
y
2 The locus of
P y
1 4
x
2 5 2
x
21 4
P. 11
23.2 Equations of Circles
A. Circles
In section 23.1, we learnt that the locus of a moving point that keeps a fixed distance from a fixed point is a circle.
Note:
The fixed distance is the radius and the fixed point is the centre of the circle. In this section, we put the circle on the coordinate plane and hence find the equation of the circle.
P. 12
23.2 Equations of Circles
A. Circles
Suppose the centre is
C
(
h
,
k
) and the radius is
r
. Let
P
(
x
,
y
) be the moving point on the circle, then from the distance formula, we have (
x
h
) 2 (
y
k
) 2
r
.
Taking square on both sides, we have (
x
h
) 2 + (
y
k
) 2
r
2 . Thus, the equation of the circle is (
x
h
) 2 where centre ( + (
h
,
y k
k
) 2
r
2 , ) and radius
r
.
If the centre is the origin (0, 0), then the equation of the circle is
x
2 +
y
2
r
2 .
Note:
This equation is called the
centre-radius form
of the circle.
P. 13
23.2 Equations of Circles
B. General Form of Equations of Circles
If we expand the equation of the circle (
x
h
) 2 + (
y
k
) 2
r
2 , the equation can be expressed in the form
x
2 +
y
2
x
2 2 2
hx hx
+
h
2 2
ky
+
y
2 + (
h
2 + 2
ky k
2 +
k
2
r
2 )
r
2 0 i.e.,
x
2 where +
y
2
D
+
Dx
2
h
,
E
+
Ey
+
F
2
k
and 0,
F
h
2 +
k
2
r
2 . This is called the
general form
of the equation of a circle.
Notes:
1. The coefficients of
x
2 and
y
2 are both equal to 1.
2.
D
,
E
and
F
can be any real numbers.
3. The right hand side of the general form of the equation of a circle is 0.
P. 14
23.2 Equations of Circles
B. General Form of Equations of Circles
Example 23.7T
If a circle is centred at ( 4, 2) and it passes through ( 2, 2), find the equation of the circle. Express the answer in the general form.
Solution:
The radius of the circle 2 ( 4) 2.
The equation of the circle:
x
2 8
x
(
x
4 ) 2 16
y
(
y
2 ) 2 2 4
y
4 2 2 4
x
2
y
2 8
x
4
y
16 0
P. 15
23.2 Equations of Circles
B. General Form of Equations of Circles
Example 23.8T
Given that
A
(3, 5) and
B
( 9, 7) are the two end points of a diameter of a circle. (a) Find the equation of the circle in general form. (b) Find the coordinates of the points of intersection of the circle and the
x
-axis.
Solution:
(a) Centre Radius 3 9 2 ( 3 , 6 , 5 ) 1 2 ( 3 7 2 9 ) 2 ( 5 37 The equation of the circle: (
x
2 ( 2
x
2 6
x
9 3 )
y
2 12
y y
6 ) 36
x
2
y
2 6
x
12
y
8 7 ) 2 ( 37 0 (b) When
y
(
x
x
2 0, we have 2 )( 6
x x
4 8
x
) 0 0 4 or 2 The points of intersection 37 ) 2 of the circle and the are ( 4, 0) and (
x
-axis 2, 0).
P. 16
23.2 Equations of Circles
B. General Form of Equations of Circles
Example 23.9T
A circle passes through three points
O
(0, 0),
A
(6, 0) and
B
(0, 10). (a) Find the equation of the circle in the general form. (b) Does the point
I
( 1, 2) lie on the circle?
Solution:
(a) Let the equation of the circle be
x
2 +
y
2 +
Dx
+
Ey
+
F
0 …(*) Since the three points
O
,
A
and
B
must satisfy (*), 0 6 0 2 2 2 0 0 2 2
D
( 0 )
D
( 6 ) ( 10 ) 2
E
( 0 )
F E
( 0 )
F D
( 0 ) 0 0
E
( 10 )
F
0 36 100 6 10
D E
F F F
0 ..........
( 1 ) 0 ..........
( 2 ) 0 ..........
( 3 )
P. 17
23.2 Equations of Circles
B. General Form of Equations of Circles
Example 23.9T
A circle passes through three points
O
(0, 0),
A
(6, 0) and
B
(0, 10). (a) Find the equation of the circle in the general form. (b) Does the point
I
( 1, 2) lie on the circle?
Solution:
Substituting (1) into (2) and (3), 36 100 6
D
10
E
0 ..........
( 4 0 ..........
( 5 ) ) From (4), 6
D D
36 6 From (5), 10
E E
100 10 The equation of the circle is
x
2 +
y
2 – 6
x
+ 10
y
0.
P. 18
23.2 Equations of Circles
B. General Form of Equations of Circles
Example 23.9T
A circle passes through three points
O
(0, 0),
A
(6, 0) and
B
(0, 10). (a) Find the equation of the circle in the general form. (b) Does the point
I
( 1, 2) lie on the circle?
Solution:
(b) Substituting ( 1, 2) into the equation
x
2 +
y
2 – 6
x
+ 10
y
0.
∵ L.H.S. ( 1) 2 + ( 2) 2 9 R.H.S. 6( 1) + 10( 2) ( 1, 2) does not satisfy the equation of the circle.
I
( 1, 2) does not lie on the circle.
P. 19
23.2 Equations of Circles
B. General Form of Equations of Circles
Example 23.10T
The centre of a circle lies on the straight line
x
passes through
S
(1, 3) and
T
( 2, 0). + 2
y
+ 1 0. The circle (a) Find the equation of the circle in the general form. (b) If
PS
is a diameter of the circle, find the coordinates of
P
.
Solution:
(a) Let
R
(
h
,
k
) be the centre of the circle. Since the centre lies on
x h
+ 2
k
+ 1 + 2
y
+ 1 0…………(1) 0, we have (
h
1 ) 2 (
k
3 ) 2 (
h
1 ) 2 (
k
3 ) 2 (
h
(
h
2 ) 2 2 ) 2 (
k
0 ) 2
k
2
h
2 2
h
1
k
2 6
k
9 2
h
6
h
6
k
6
k
10 6
h
k
1
h
2 4
h
0 4
h
4 4
k
2 0 ..........
..( 2 )
P. 20
23.2 Equations of Circles
B. General Form of Equations of Circles
Example 23.10T
The centre of a circle lies on the straight line
x
passes through
S
(1, 3) and
T
( 2, 0). + 2
y
+ 1 0. The circle (a) Find the equation of the circle in the general form. (b) If
PS
is a diameter of the circle, find the coordinates of
P
.
Solution:
(1) 2 (2):
k
k
1 ( 1 ) 0 Substituting
k k
2
k
0 2 2 into (2), we have
h
2 1 0
h
The centre 3 (3, 2) Radius ( 3 1 ) 2 ( 2 3 ) 2 29 The equation of the circle:
x
2
x
6 (
x
3 ) 2
x
9
y
( 2
y
4
y
2 ) 2 4 ( 29 29 ) 2 2
y
2 6
x
4
y
16 0
P. 21
23.2 Equations of Circles
B. General Form of Equations of Circles
Example 23.10T
The centre of a circle lies on the straight line
x
passes through
S
(1, 3) and
T
( 2, 0). + 2
y
+ 1 0. The circle (a) Find the equation of the circle in the general form. (b) If
PS
is a diameter of the circle, find the coordinates of
P
.
Solution:
(b) Let (
x
,
y
) be the coordinates of
P
. Since (3, 2) is the mid-point of
P
x y
2
x
2 1 3 5 3 2 and
S
, we have
y
7 The coordinates of
P
are (5, 7).
P. 22
23.2 Equations of Circles
C. Features of Equations of Circles
For the equation of a circle
x
2 +
y
2 +
Dx
+
Ey
+
F
0, we have (a) centre
D
, 2
E
2 ; (b) radius
D
2 2
E
2 2
F
Remarks:
2 2 1. If
D
E
F
> 0, then the radius is a real number. 2 2 This kind of circle is known as a
real circle
. 2. If
D
2 2 zero radius.
E
2 2
F
0, then the equation represents a circle with This kind of circle is known as a
point circle
. 2 2 3. If
D
E
F
< 0, then the radius is not a real number. 2 2 This kind of circle is known as an
imaginary circle
.
P. 23
23.2 Equations of Circles
C. Features of Equations of Circles
Example 23.11T
Consider a circle:
x
2 +
y
2 + 5
x
– 10
y
+ 15 0 Determine whether the point
A
(1, –1) lies on, inside or outside the circle.
Solution:
Centre 5 2 , ( 10 ) 2 5 2 , 5 Radius 5 2 2 ( 5 ) 2 15 65 4 Distance between the point
A
and the centre 5 2 1 2 [ 5 ( 1 )] 2 193 4 radius
A
(1, 1) lies outside the circle.
P. 24
23.2 Equations of Circles
C. Features of Equations of Circles
Example 23.12T
The following are the equations of two circles:
C
1 :
x
2 +
y
2 – 2
x
– 14
y
+ 46
C
2 :
x
2 +
y
2 – 12
x
+ 10
y
0 – 164 0 (a) Find the centres and radii of the two circles. (b) Hence show that the two circles touch each other internally.
Solution:
(a) For
C
1 , centre ( 2 ) , 2 ( 14 ) 2 ( 1 , 7 ) and radius ( 1 ) 2 ( 7 ) 2 46 2 For
C
2 , centre radius ( 12 ) 2 , 10 2 ( 6 , 5 ) ( 6 ) 2 5 2 ( 164 ) 15 and
P. 25
23.2 Equations of Circles
C. Features of Equations of Circles
Example 23.12T
The following are the equations of two circles:
C
1 :
x
2 +
y
2 – 2
x
– 14
y
+ 46
C
2 :
x
2 +
y
2 – 12
x
+ 10
y
0 – 164 0 (a) Find the centres and radii of the two circles. (b) Hence show that the two circles touch each other internally.
Solution:
(b) Distance between the centres ( 1 6 ) 2 ( 7 13 169 Difference of the radii of the two circles 15 – 2 13 Distance between the centres The two circles touch each other internally. 5 ) 2
P. 26
In the same coordinate plane, there are three cases showing the relationship between the graphs of a circle and a straight line
y
mx
+
c
.
x
2 +
y
2 +
Dx
+
Ey
+
F
0
Case 1 :
intersect at two distinct points;
Case 2 :
intersect at one point only;
Case 3 :
no point of intersection.
Notes:
For case 2, the straight line is called a tangent to the circle.
P. 27
Without the actual drawing of the graphs, the number of points of intersection of the two graphs can be determined algebraically by carrying out the following steps.
Step 1:
Use the method of substitution to eliminate one of the unknowns (either
x
or
y
) of the simultaneous equations. We can then obtain a quadratic equation in one unknown.
y x
2
mx y
2
c
..........
Dx
..........
Ey
F
..........
0 .( .....( 1 ) 2 ) Substituting (1) into (2),
x
(1 + 2 +
m
2
m
2 )
x
2
x
2 + (
mx
+
c
) 2
x
2 + 2
mcx
+
c
2 + +
Dx Dx
+ +
E
(
mx Emx
+
c
) + +
Ec
+
F F
+ (2
mc
+
D
+
Em
)
x
+ (
c
2 +
Ec
+
F
) 0 0 0 ... (*)
Step 2:
Evaluate the discriminant ( D ) of the quadratic equation (*). If If D If D > 0, there are two points of intersection. D 0, there is only one point of intersection. < 0, there is no point of intersection.
P. 28
Example 23.13T
Find the coordinates of the points of intersection of the straight line 2
x
–
y
– 1 0 and the circle
x
2 +
y
2 – 3
x
+
y
– 10 0.
Solution:
Consider the simultaneous equations of the straight line and the circle.
y x
2 2
y x
2 1 ..........
3
x
y
..........
10 ..........
0 ......( ..........
...( 1 ) 2 ) Substituting (1) into (2), we have
x
2
x
2 ( 4
x
2 2
x
1 ) 4
x
2 1 3
x
3
x
( 2
x
2
x
1 ) 1 10 10 5
x
2 (
x x
5
x
10 2 2 )(
x x
2 1 )
x
0 0 0 0 0 1 or 2 ( When
y
x
1, 2( 1) 1 3 When
y
1,
x
2(2) 2, 1 The points of 3 intersection are 3) and (2, 3).
P. 29
Example 23.14T
If the straight line
y
– 3
x
– 1 0 meets the circle
x
2 +
y
2 + 2
x
+ 4
y
+
k
0 at two distinct points, find the range of values of
k
.
Solution:
Consider the simultaneous equations of the straight line and the circle.
y x
2 3
x y
2 1 ..........
2
x
4 ..........
y
k
..........
0 .....( ..........
.( 1 ) 2 ) Substituting (1) into (2),
x
2 ( 3
x
1 ) 2 2
x
x
2 9
x
2 6
x
1 10
x
2 2
x
4 ( 12
x
20 3
x x
1 )
k
4
k
( 5
k
) 0 0 0 ........( ) Since the straight line meets the circle at two distinct points, the discriminant of (*) is greater than 0.
20 2 4 ( 10 )( 5
k
) 0 400 40 ( 5
k
) 10
k
5 5
k
The range is
k
< 5.
P. 30
Example 23.15T
Given a circle (
x
– 1) 2 + (
y
+ 2) 2
k
. If
P
(4, –3) lies on the circle, find (a) the centre of the circle, (b) the value of
k
, (c) the equation of the tangent to the circle at
P
.
Solution:
(a) Centre ( 1 , 2 ) (b) Substituting (4, 3) into the equation (
x
1 ) 2 ( 4 1 ) 2 ( 3 9 2 ) 2 1
k
k k
10 (
y
2 ) 2
k
,
P. 31
Example 23.15T
Given a circle (
x
– 1) 2 + (
y
+ 2) 2
k
. If
P
(4, –3) lies on the circle, find (a) the centre of the circle, (b) the value of
k
, (c) the equation of the tangent to the circle at
P
.
Solution:
(c) Slope of the line joining the centre and
P
Slope of the tangent 1 1 3 3 2 1 ( 3 ) 4 1 3 Equation of the tangent: 3
x y
y
(
y
3 ) 3 15 3 0 ( 3
x x
4 12 )
P. 32
Chapter Summary
23.1 Concept of Loci
1. A locus is the set of all points that satisfy the given specified conditions.
2. The locus of points can be expressed in algebraic equations.
P. 33
Chapter Summary
23.2 Equations of Circles
1. A circle can be expressed by the centre-radius form (
x
h
) 2 + (
y
k
) 2
r
2 , where centre (
h
,
k
) and radius
r
. 2. The general form of a circle is
x
2 +
y
2 +
Dx
+
Ey
+
F
0. 3. From the general form, we have (a) centre (b) radius
D
, 2
E
2
D
2 2
E
2 and 2
F
4. If the distance between the centre of a circle and a point
P
is (a) smaller than the radius of the circle, then
P
lies inside the circle; (b) equal to the radius of the circle, then
P
lies on the circle; (c) greater than the radius of the circle, then
P
lies outside the circle.
P. 34
Chapter Summary
23.3 Intersection of a Straight Line and a Circle
1. There are three cases showing the relationship between a straight line and a circle: (a) Intersect at two points (b) Intersect at one point (c) No point of intersection We can use the discriminant to determine the number of the points of intersection.
2. The coordinates of the point(s) of intersection can be found by solving the simultaneous equations of a straight line and a circle.
3. If a straight line touches the circle at one point only, then the straight line is a tangent to the circle.
P. 35