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23. Locus. Case Study. 23.1 Concept of Loci. 23.2Equations of Circles. 23.3Intersection of a Straight Line and a Circle. Chapter Summary. The satellite should move in a way that prevents itself from deviating from the path. Do you know how the Chang’e 1 Satellite orbited the Moon?. - PowerPoint PPT Presentation

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Page 1: 23.2Equations of Circles

23

23.2 Equations of Circles

23.3 Intersection of a Straight Line and a Circle

Chapter Summary

Case Study

Locus

23.1 Concept of Loci

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P. 2

On 24 October 2007, China’s first lunar orbited ‘Chang’e 1 Satellite’ was launched from Xichuan.

Case StudyCase Study

The satellite orbited the Moon at an altitude of 200 km above the lunar surface.

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.

It carried out a one-year lunar exploration mission.

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P. 3

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.

23.1 23.1 Concept of LociConcept of Loci

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.

If a point is moving under a specific condition, then its path is called the locus of the moving point.

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.

A. Description of LociA. Description of Loci

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Locus :The locus of the point R is a pair of straight lines which are (1) parallel to L; (2) on one of the sides of L respectively; (3) equidistant from L with a distance of 3 cm.

23.1 23.1 Concept of LociConcept 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.

A. Description of LociA. Description of Loci

Locus :The locus of the point Q is a straight line which is (1) parallel to both AB and CD; (2) in the middle of AB and CD; (3) equidistant from AB and CD with a distance of 2 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.

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Example 23.1T

Solution:

Describe the locus of the moving point P under each of the following conditions. (a) P is always 3 cm from the origin O. (b) AB is a straight line and PAB PBA.

(a) A circle centred at O with radius 3 cm.

23.1 23.1 Concept of LociConcept of Loci

A. Description of LociA. Description of Loci

(b) The perpendicular bisector of AB.

P

O

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P. 6

Example 23.2TIn the figure, l1 and l2 are two parallel lines 6 cm apart. A moving point P is equidistant from l1 and l2. Sketch the locus of P.

Solution:The locus is a line which is parallel to l1 and l2, and is 3 cm from l1 and l2.

23.1 23.1 Concept of LociConcept of Loci

A. Description of LociA. Description of Loci

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P. 7

Example 23.3TIn 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.

23.1 23.1 Concept of LociConcept of Loci

A. Description of LociA. Description of Loci

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P. 8

Example 23.4T

Solution:

AB is a line segment. A moving point P passes through A and B and moves in a way such that APB 130°. Describe and sketch the locus of P.

The locus of P is a minor arc APB of a circle.

23.1 23.1 Concept of LociConcept of Loci

A. Description of LociA. Description of Loci

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P. 9

23.1 23.1 Concept of LociConcept of Loci

Besides verbal description of the locus, we can also describe the locus of a point in an algebraic equation.

BB. Descri. Describe Loci in Algebraic Equationsbe Loci in Algebraic Equations

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P. 10

Example 23.5T

Solution:

A moving point P is always units from . Express the locus of P in algebraic form.

Let (x, y) be the coordinates of P.

The locus of P is 4x2 + 4y2 – 4x + 8y – 3

0.

1,

2

1Q2

Since the distance between P and Q is ,

22

)1(2

1

yx

2)1(2

1 22

yx

2)1(2

1 22

yx

2124

1 22 yyxx

04

3222 yxyx

23.1 23.1 Concept of LociConcept of Loci

BB. Descri. Describe Loci in Algebraic Equationsbe Loci in Algebraic Equations

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P. 11

Example 23.6T

Solution:

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 ax2 + bx + c.

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)

The locus of P is .

2222 )0()5()]2([)( yxyxx222 )5()2( yxy

222 251044 yxxyy 21104 2 xxy

4

21

2

5

4

1 2 xxy

23.1 23.1 Concept of LociConcept of Loci

BB. Descri. Describe Loci in Algebraic Equationsbe Loci in Algebraic Equations

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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.

23.23.22 Equations of CirclesEquations of Circles

In this section, we put the circle on the coordinate plane and hence find the equation of the circle.

Note: The fixed distance is the radius and the fixed point is the centre of the circle.

AA. . CirclesCircles

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P. 13

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

23.23.22 Equations of CirclesEquations of Circles

Taking square on both sides, we have (x h)2 + (y k)2 r2.

AA. . CirclesCircles

.)()( 22 rkyhx

Thus, the equation of the circle is (x h)2 + (y k)2 r2,

where centre (h, k) and radius r.If the centre is the origin (0, 0), then the equation of the circle is x2 + y2 r2.

Note: This equation is called the centre-radius form of the circle.

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P. 14

If we expand the equation of the circle (x h)2 + (y k)2 r2, the equation can be expressed in the form

23.23.22 Equations of CirclesEquations of Circles

BB. . General Form of Equations of CirclesGeneral Form of Equations of Circles

This is called the general form of the equation of a circle.

i.e., x2 + y2 + Dx + Ey + F 0, where D 2h, E 2k and F h2 + k2 r2.

Notes: 1. The coefficients of x2 and y2 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.

x2 2hx + h2 + y2 2ky + k2 r2

x2 + y2 2hx 2ky + (h2 + k2 r2) 0

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Example 23.7T

Solution:

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.

The radius of the circle 2 (4) 2.

The equation of the circle: 222 2)2()4( yx

444168 22 yyxx

0164822 yxyx

23.23.22 Equations of CirclesEquations of Circles

BB. . General Form of Equations of CirclesGeneral Form of Equations of Circles

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Example 23.8T

Solution:

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.

23.23.22 Equations of CirclesEquations of Circles

BB. . General Form of Equations of CirclesGeneral Form of Equations of Circles

(a) Centre

Radius

2

75 ,

2

93

22 )75()93(2

1

The equation of the circle: 222 )37()6()3( yx

37361296 22 yyxx0812622 yxyx

37

)6 ,3(

(b) When y 0, we have

x 4 or 2 The points of intersection o

f the circle and the x-axis are (4, 0) and (2, 0).

0862 xx0)4)(2( xx

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Example 23.9T

Solution:

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?

(a) Let the equation of the circle be x2 + y2 + Dx + Ey + F 0 …(*)

Since the three points O, A and B must satisfy (*),

0)10()0()10(0

0)0()6(06

0)0()0(00

22

22

22

FED

FED

FED

)3(..........010100)2(..........0636)1(..........0

FEFDF

23.23.22 Equations of CirclesEquations of Circles

BB. . General Form of Equations of CirclesGeneral Form of Equations of Circles

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Example 23.9T

Solution:

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?

Substituting (1) into (2) and (3),

From (4), 6D 36D 6

)5(..........010100 )4(..........0636

ED

From (5), 10E 100 E 10

The equation of the circle is x2 + y2 – 6x + 10y 0.

23.23.22 Equations of CirclesEquations of Circles

BB. . General Form of Equations of CirclesGeneral Form of Equations of Circles

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Example 23.9T

Solution:

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?

(b) Substituting (1, 2) into the equation x2 + y2 – 6x + 10y 0.

L.H.S. (1)2 + (2)2 6(1) + 10(2) 9 R.H.S.

∵ (1, 2) does not satisfy the equation of the circle.

I(1, 2) does not lie on the circle.

23.23.22 Equations of CirclesEquations of Circles

BB. . General Form of Equations of CirclesGeneral Form of Equations of Circles

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Solution:

The centre of a circle lies on the straight line x + 2y + 1 0. The circle passes through S(1, 3) and T(2, 0).  (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.

(a) Let R(h, k) be the centre of the circle. Since the centre lies on x + 2y + 1 0, we have

h + 2k + 1 0…………(1)2222 )0()2()3()1( khkh

2222 )2()3()1( khkh 2222 449612 khhkkhh

441062 hkh

)2..(..........01 kh0666 kh

23.23.22 Equations of CirclesEquations of Circles

BB. . General Form of Equations of CirclesGeneral Form of Equations of Circles

Example 23.10T

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Solution:

The centre of a circle lies on the straight line x + 2y + 1 0. The circle passes through S(1, 3) and T(2, 0).  (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.

(1) (2): 0)1(12 kk

202

k

k

Substituting k 2 into (2), we have

3012

hh

The centre (3, 2)

Radius 22 )32()13(

The equation of the circle:222 )29()2()3( yx

294496 22 yyxx

0164622 yxyx

23.23.22 Equations of CirclesEquations of Circles

BB. . General Form of Equations of CirclesGeneral Form of Equations of Circles

Example 23.10T

29

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P. 22

Solution:

The centre of a circle lies on the straight line x + 2y + 1 0. The circle passes through S(1, 3) and T(2, 0).  (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.

(b) Let (x, y) be the coordinates of P.

Since (3, 2) is the mid-point of P and S, we have

22

3

32

1

y

x

7

5

y

x

The coordinates of P are (5, 7).

23.23.22 Equations of CirclesEquations of Circles

BB. . General Form of Equations of CirclesGeneral Form of Equations of Circles

Example 23.10T

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23.23.22 Equations of CirclesEquations of Circles

CC. . Features of Equations of CirclesFeatures of Equations of Circles

Remarks:

For the equation of a circle x2 + y2 + Dx + Ey + F 0, we have

(a) centre (b) radius;

2,

2

ED

FED

22

22

1. If > 0, then the radius is a real number. FED

22

22This kind of circle is known as a real circle.

2. If 0, then the equation represents a circle with

zero radius.

FED

22

22

This kind of circle is known as a point circle.

3. If < 0, then the radius is not a real number. FED

22

22This kind of circle is known as an imaginary circle.

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P. 24

Solution:

Consider a circle: x2 + y2 + 5x – 10y + 15 0 Determine whether the point A(1, –1) lies on, inside or outside the circle.

Centre

Distance between the point A and the centre

23.23.22 Equations of CirclesEquations of Circles

CC. . Features of Equations of CirclesFeatures of Equations of Circles

5 ,

2

5

2

)10( ,

2

5

Radius 4

6515)5(

2

5 22

22

)]1(5[12

5

4

193 radius

A(1, 1) lies outside the circle.

Example 23.11T

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P. 25

Solution:

The following are the equations of two circles: C1 : x2 + y2 – 2x – 14y + 46 0 C2 : x2 + y2 – 12x + 10y – 164 0

(a) Find the centres and radii of the two circles. (b) Hence show that the two circles touch each other internally.

(a) For C1, centre and

23.23.22 Equations of CirclesEquations of Circles

CC. . Features of Equations of CirclesFeatures of Equations of Circles

)7 ,1(2

)14( ,

2

)2(

radius 246)7()1( 22

For C2, centre and

radius

)5 ,6(2

10 ,

2

)12(

15)164(5)6( 22

Example 23.12T

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P. 26

Solution:

(b) Distance between the centres

23.23.22 Equations of CirclesEquations of Circles

CC. . Features of Equations of CirclesFeatures of Equations of Circles

Difference of the radii of the two circles 15 – 2 13 Distance between the centres

22 )57()61(

13169

The two circles touch each other internally.

Example 23.12TThe following are the equations of two circles:

C1 : x2 + y2 – 2x – 14y + 46 0 C2 : x2 + y2 – 12x + 10y – 164 0

(a) Find the centres and radii of the two circles. (b) Hence show that the two circles touch each other internally.

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In the same coordinate plane, there are three cases showing the relationship between the graphs of a circle x2 + y2 + Dx + Ey + F 0 and a straight line y mx + c.

23.23.33 Intersection of a Straight Line Intersection of a Straight Line and a Circleand a Circle

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.

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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.

23.23.33 Intersection of a Straight Line Intersection of a Straight Line and a Circleand a Circle

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.

)2.....(0)1.(..............................

22 FEyDxyxcmxy

Substituting (1) into (2), x2 + (mx + c)2 + Dx + E(mx + c) + F 0

x2 + m2 x2 + 2mcx + c2 + Dx + Emx + Ec + F 0 (1 + m2 )x2 + (2mc + D + Em)x + (c2 + Ec + F) 0 ... (*)

Step 2: Evaluate the discriminant () of the quadratic equation (*). If > 0, there are two points of intersection. If 0, there is only one point of intersection. If < 0, there is no point of intersection.

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Solution:

Find the coordinates of the points of intersection of the straight line 2x – y – 1 0 and the circle x2 + y2 – 3x + y – 10 0.

Consider the simultaneous equations of the straight line and the circle.

Substituting (1) into (2), we haveWhen x 1, y 2(1) 1 3

)2...(..........0103)1......(..............................12

22 yxyxxy

010)12(3)12( 22 xxxx010123144 22 xxxxx When x 2,

y 2(2) 1 3

The points of intersection are (1, 3) and (2, 3).

23.23.33 Intersection of a Straight Line Intersection of a Straight Line and a Circleand a Circle

Example 23.13T

x 1 or 2

01055 2 xx022 xx0)1)(2( xx

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Solution:

If the straight line y – 3x – 1 0 meets the circle x2 + y2 + 2x + 4y + k 0 at two distinct points, find the range of values of k.

Consider the simultaneous equations of the straight line and the circle.

Substituting (1) into (2),

Since the straight line meets the circle at two distinct points, the discriminant of (*) is greater than 0.

The range is k < 5.

)2.(..........042)1.....(..............................13

22 kyxyxxy

0)13(42)13( 22 kxxxx

04122169 22 kxxxxx

)........(0)5(2010 2 kxx

0)5)(10(4202 k)5(40400 k

k 5105k

23.23.33 Intersection of a Straight Line Intersection of a Straight Line and a Circleand a Circle

Example 23.14T

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P. 31

Solution:

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.

(a) Centre

(b) Substituting (4, 3) into the equation

)2 ,1(

,)2()1( 22 kyx k 22 )23()14(k1910k

23.23.33 Intersection of a Straight Line Intersection of a Straight Line and a Circleand a Circle

Example 23.15T

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P. 32

Solution:

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.

(c) Slope of the line joining the centre and P

Slope of the tangent

41

)3(2

3

1

3

11 3

Equation of the tangent:)4(3)3( xy

1233 xy0153 yx

23.23.33 Intersection of a Straight Line Intersection of a Straight Line and a Circleand a Circle

Example 23.15T

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23.1 Concept of Loci

1. A locus is the set of all points that satisfy the given specified conditions.

Chapter Chapter SummarySummary

2. The locus of points can be expressed in algebraic equations.

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1. A circle can be expressed by the centre-radius form (x h)2 + (y k)2 r2, where centre (h, k) and radius r.

Chapter Chapter SummarySummary23.2 Equations of Circles

2. The general form of a circle is x2 + y2 + Dx + Ey + F 0.

3. From the general form, we have

(a) centre and

(b) radius

2,

2

ED

FED

22

22

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.

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Chapter Chapter SummarySummary23.3 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 intersectionWe 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.