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# Mathematics SL portoflio type 1(circles)

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Introduction

Background information:

The general form of a circle is denoted by where r is the radius of the circle, k stands for the vertical translation and h for the horizontal translation of the circle on the coordinate plane. We derive the equation of the circle by transforming the general formula to get , where + sign corresponds to the upper half of the circle and the – sign corresponds to the bottom half of the circle. This equation will be used throughout the investigation to assist with algebraic problems concerning the relationship between r, OP and OP’.  For the purpose of this investigation, O will correspond to (0,0) on the coordinate plane.

Part 1

The purpose of this exercise is to determine the relationship between OP and OP’ with constant r of 1. However, it is important to note that r only implies to the radius of circles C1 and C3.

Equation of circle C2   Equation of circle C1   Centre of circle C3 is the upper intersection of circles C1 and C2 , thus we can find the middle of circle C3 by determining the value of x and y where both formula of the circles C1 and C2  are equal to each other, as shown below. We take the positive value of the root of the equation, as we are interested in the upper intersection of the two circles, which corresponds to the circle C1.

C2=C1      ;      The coordinate of midpoint of circle C3 is Middle       Similarly, the same technique was used to find OP’ with different values of r and constant value of OP of 2

The results have been displayed in the table below:

 r 2 3 4 OP’ 2 4.5 8 When the value of OP is kept constant at 2, OP’ is equivalent to r2 divided by 2. However, there are some limitations to the equation above. By studying the diagram below we can see that as the radius increases, the value of OP’ also increases. My statement is supported by the data in the table above and on the observations of the diagrams below. However, if the value of r is beyond a certain point, there will be no intersection between circles C1 and C2, thus giving no value point A and hence OP’ (the circle C3 will not exist). The maximum value of r (with constant OP of 2) which gives the solution for OP’ is 4. Radius above 4 will lead to the circle C2 to be inside the circle C1 hence giving no intersection between the 2 circles and no solution for OP’. On the diagram below, when can clearly see the situation when C2 is inside C1, with not intersection of the two circles, leading to the non-existence of C3 and giving no solution for OP’

From the diagrams below, we can see that for constant OP of two, increasing the radius casues the value of OP’ to increase ,untill r reaches its maximum to give a solution .   Maximum value of OP (4)

Conclusion

1 and C2, hence, point A will not exist and there is no solution for OP’. However, not always when r is bigger than OP, there is no solution for OP’. This has been further investigated in Part 2 using graphing technology.

Conclusion

Throughout the investigating we have found general statement which link all three variables: OP (radius of C2 or distance between O and C2) , r (radius of C1 and C3) and OP’(the intersection of C3 with the x-axis or OP.  The formula is represented by: Through part A, we have found the relation between OP and OP’, with a constant r. In part B we have found the relation of OP’ and r, with constant OP and in C we have found the combined general formula by using the data generated in part A and B, and manipulating all variables r, OP and OP’.

Limitations of the general formula were determined by the use of technology, which would give no values of OP’,(through the non-existence of circle C3). r could not be greater than OP as this would result in no intersection between C1 and C2 and would not lead to solution for OP’. For this reason, r must be smaller or equal to OP for there to be a value of OP’ and for the combined general statement to exist.

Appendix A (for part 3)      OP = 2           r=1         OP’=0.5 When OP=3; r=1 then OP’= 1/3   When OP=-2; r=1 then OP’=-1/2

Due to inaccuracy of the results obtained using graphing technology, we assume that the P’ is (- 1/3, 0) When OP=-3; r=1 then OP’=- 1/3

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

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