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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 with centre O and radius r, and any point P.

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

The circle has centre P and radius . A is one of the points of intersection of and . Circle  has centre A, and radius r. The point P’ is the intersection of with (OP). The r=1, =2, and P’=0.5. This is shown in the diagram below.

This investigation will explore in depth of the relationship between the r value and  values, when r is held constant and  values modified. It will also investigate the reverse, the relationship when the   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   when =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.

In ΔAOP’, lines  and  have the same length, because both points, O and P’ are within the circumference of the circle, which means that  and   are its radius. Similarly, ΔAOP forms another isosceles triangle, because the lines   and   are both radii of the circle .

=

Middle

1, smaller values were also tried.

In this example, the length of =  , r= 1. Calculations resulted that  still equaled to . The following graph displayed is when r=1, =  , and =  = 1.

In the next example, the length of = , r= 1. Calculations resulted that  still equaled to . The following graph displayed is when r= 1, = , and =  = 1..

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

The following graph when = , r= 1.

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

Limitations in the general statement,  , (n= ) is that 0.5. The value 0.5 will only work for the case when r=1 and values are changed. So in a broader sense, when radius is held constant, limitations to | ,   , rN, natural number｝. The (this is the radius of )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  when the length of  is held constant and the radius is changed.

The circle has centre P and radius . A is one of the points of intersection of and . Circle  has centre A, and radius r. The point P’ is the intersection of with (OP). The r=2, =2.

Conclusion

r |< rN, 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:

, (n= )

and

’ =  (n= r).

The first general statement is consistent with the second general statement. It is why when you calculate the lengths of , 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  and  is held constant, with changing radius of  (), while the second general statement is when the radius of  and are modified, while the radius of  () remains constant.

Conclusion

 r . ’ 1 Not defined 1 2 1 1 1 1 1 1 1 2 1 3 1 4 1 5 1 100 1 1000

and (when r is held constant) are inversely related (). Therefore, the general statement to represent this would be:

, (n= )

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

 r . ’ 2 Not defined 2 Not defined 2 1 2 2 2 = 2 3 2 4 2 = 8 5 2 Not defined

and (when  is held constant) is an exponential relationship (). Therefore, the general statement to represent this would be:

’ =  (n= r).

Limitations in this general statement would be r |< rN, Natural Numbers｝. If the radiusis greater than length of ’s squared value, then there wouldn’t be any point of intersection between  and . When the radius is less than the inverse of , and  do not intersect.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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