- Level: International Baccalaureate
- Subject: Maths
- Word count: 2118
Math SL Circle Portfolio. The aim of this task is to investigate positions of points in intersecting circles.
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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
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 ｛| ≤ , , r∈ N, 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
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 ｛| ≤ , r∈ N, 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 |< r≤N, 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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