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Math SL Circle Portfolio. The aim of this task is to investigate positions of points in intersecting circles.

Extracts from this document...

Introduction

Math IA (SL TYPE I)                                                                    

Circles

Jennifer

Aim: The aim of this task is to investigate positions of points in intersecting circles.

Introduction

The following diagram shows a circle image00.pngimage00.pngwith centre O and radius r, and any point P.

image39.png

In this case, the r is the distance between any point (such as A) and the centre point O of the circle image00.pngimage00.png. Since the radius is 1 unit, OP will also be 1 unit away from the point O.

The circleimage06.pngimage06.png has centre P and radius image24.png. A is one of the points of intersection of image00.pngimage00.pngand image06.pngimage06.png. Circle image09.pngimage09.png has centre A, and radius r. The point P’ is the intersection of image09.pngimage09.pngwith (OP). The r=1, image24.pngimage24.png=2, and P’=0.5. This is shown in the diagram below.

image34.png

This investigation will explore in depth of the relationship between the r value and image40.pngimage40.png values, when r is held constant and image24.pngimage24.png values modified. It will also investigate the reverse, the relationship when the image24.pngimage24.png  values are held constant and the r values are modified.

In the first example, the r value given is 1. An analytic approach will be taken to find the length of image24.pngimage24.png  when image24.pngimage24.png=2. Firstly, one can note that 2 isosceles triangle can be drawn by using the points A, O, P’, and P. It is shown in the diagram below.

image57.png

In ΔAOP’, lines image58.pngimage58.png and image63.pngimage63.png have the same length, because both points, O and P’ are within the circumference of the circleimage09.pngimage09.png, which means that image58.pngimage58.png and  image63.pngimage63.png are its radius. Similarly, ΔAOP forms another isosceles triangle, because the lines  image63.pngimage63.png and  image24.pngimage24.png are both radii of the circle image06.pngimage06.png.

image58.pngimage58.png=

...read more.

Middle

 1, smaller values were also tried.

In this example, the length of image24.pngimage24.png= image44.pngimage44.png , r= 1. Calculations resulted that image45.pngimage45.png still equaled to image46.pngimage46.png. The following graph displayed is when r=1, image24.pngimage24.png= image44.pngimage44.png , and image45.pngimage45.png= image46.pngimage46.png = 1.image42.pngimage42.png

image47.png

In the next example, the length of image24.pngimage24.png= image48.pngimage48.png, r= 1. Calculations resulted that image45.pngimage45.png still equaled to image49.png. The following graph displayed is when r= 1, image24.pngimage24.png= image48.pngimage48.png, and image45.pngimage45.png= image49.pngimage49.png = 1.image50.pngimage50.png.

image51.png

In the following graph, image24.pngimage24.png= image52.pngimage52.png , r= 1. Through the same method done by the sample calculation, the image24.pngimage24.png= 2, which makes the general statement still valid.

image53.png

The following graph when image24.pngimage24.png= image54.pngimage54.png, r= 1.

image55.png

In the foreshown graph, circle image09.pngimage09.png cannot be drawn, because there is no intersection point (point A) between image00.pngimage00.pngandimage06.pngimage06.png. This makes sense, because if the radius of image06.pngimage06.pngis less than half the radius of image00.pngimage00.png, (in which this case is 0.5), then the circle won’t be large enough to intersect with image00.pngimage00.png.

Limitations in the general statement, image30.pngimage30.png , (n= image24.pngimage24.png) is that image24.pngimage24.png0.5. The value 0.5 will only work for the case when r=1 and image24.pngimage24.pngvalues are changed. So in a broader sense, when radius is held constant, limitations to image24.pngimage24.png| image24.pngimage24.pngimage56.png,   image24.pngimage24.png, rN, natural number}. The image24.pngimage24.png(this is the radius of image06.pngimage06.png)andradius must be a natural number, because it cannot be 0, or be a negative number, because it refers to length.

The second part of the investigation will focus on finding a relationship of the length of image45.pngimage45.png when the length of image24.pngimage24.png is held constant and the radius is changed.

The circleimage06.pngimage06.png has centre P and radius image24.pngimage24.png. A is one of the points of intersection of image00.pngimage00.pngand image06.pngimage06.png. Circle image09.pngimage09.png has centre A, and radius r. The point P’ is the intersection of image09.pngimage09.pngwith (OP). The r=2, image24.pngimage24.png=2.

...read more.

Conclusion

 r |image107.pngimage107.png< rimage108.pngimage108.pngimage109.pngimage109.pngN, Natural NumbersThe radius must be a natural number, because the length of something cannot be 0 or have a negative value, for it is a measurement.

The two general statements derived from this investigation are:

image38.pngimage38.png , (n= image24.png)

and

image24.pngimage24.png’ = image95.pngimage95.png (n= r).

The first general statement is consistent with the second general statement. It is why when you calculate the lengths of image24.pngimage24.png, you can use the same method for both situations. However, the derived general statement appears different, because each one is dealing with a different constant. The first general statement is when the radius of image00.pngimage00.png and image06.pngimage06.png is held constant, with changing radius of image09.pngimage09.png (image24.pngimage24.png), while the second general statement is when the radius of image00.pngimage00.png and image06.pngimage06.pngare modified, while the radius of image09.pngimage09.png (image24.pngimage24.png) remains constant.

Conclusion

r

image24.pngimage24.png.

image24.pngimage24.png

1

image110.png

Not defined

1

image26.png

2

1

image111.png

image112.png

1

image113.png

image114.png

1

image115.png

image116.png

1

1

1

1

2

image26.png

1

3

image27.png

1

4

image28.png

1

5

image117.png

1

100

image118.png

1

1000

image119.png

image24.pngimage24.pngand image21.pngimage21.png(when r is held constant) are inversely related (image21.pngimage21.pngimage29.pngimage29.png). Therefore, the general statement to represent this would be:

image30.pngimage30.png , (n= image24.pngimage24.png)

Limitations in this general statement, when radius is held constant, is image24.pngimage24.png| image24.pngimage56.pngimage56.png, rN, natural number}. If the radius of image06.pngimage06.pngis less than half the radius of image00.pngimage00.png, then the circle won’t be large enough to intersect with image00.pngimage00.png. The radius must be a natural number, because it cannot be 0, or be a negative number, because it refers to length.

r

image24.pngimage24.png.

image24.pngimage24.png

image110.png

2

Not defined

image26.png

2

Not defined

image120.png

2

image121.png

1

2

image26.png

2

2

image91.pngimage91.png= 2

3

2

image92.png

4

2

image93.pngimage93.png= 8

5

2

Not defined

image24.pngimage24.pngand image21.pngimage21.png(when image24.pngimage24.png is held constant) is an exponential relationship (image24.pngimage24.pngimage94.pngimage94.png). Therefore, the general statement to represent this would be:

image24.pngimage24.png’ = image95.pngimage95.png (n= r).

Limitations in this general statement would be r |image107.pngimage107.png< rimage108.pngimage108.pngimage109.pngN, Natural Numbers}. If the radiusis greater than length of image24.pngimage24.png’s squared value, then there wouldn’t be any point of intersection between image00.pngimage00.png and image06.pngimage06.png. When the radius is less than the inverse of image24.pngimage24.png, image09.pngimage09.pngand image24.pngimage24.png do not intersect.

...read more.

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