Because the triangle AOP’ is an isosceles triangle, with two points at the x-axis, cutting it into 2 equal pieces would produce 2 right angle triangles with equal bases of 0.25. Thus, the x-coordinate of point A (
is the midpoint of OP’, which is clearly shown on the diagram above. The distance of O to P’ is twice the distance of O to M, hence multiplying the x-coordinate of M would give us;
∴
I have used similar approach to find OP’ with OP values being 3 and 4, with constant r of 1. The results of the process have been displayed below.
Therefore we can state that when r is kept constant at value 1, OP’ is equivalent to the inverse of OP.
Part 2
The purpose of Part 2 is to determine the relationship between OP’ and r, with constant OP of 2. Again, it is important to note that r only implies to the radius of circles C1 and C3.
Finding equation for C2:
When r=2
Equation of circle C1;
Finding upper intersection of C1 and C2 or the midpoint of C3;
=
=
=
=
=
When
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:
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) giving a solution for OP’. Here OP =OP’ :
Part 3
From part A we have found the general rule for constant r value was OP’=OP-1
1÷ OP, with constant r of 1. In part B, we have found the general rule
, with constant OP of 2. Using technology the variables OP and r were changed to investigate their effect on OP’. The results have been displayed in the tables below.
Calculation of OP’ using graphing technologies have been shown in Appendix below.
The results can be merged together to give one table to increase the ease of reading, as shown below.
From the table we clearly can see that when
.
Similarly, when
and when
.
Thus, we can state that the combined general formula for all variables: OP’, OP and r is;
This general formula is also consistent for the general formula derived in part B, which was
, where the denominator 2 relates to the value of OP, which was kept constant throughout the investigation in part 2.
However, there are some limitations to the general statement. As we have observed, the OP’ does not exist for OP value of 0.3 and r value of 1,2 and 3. Further limitations will be investigated in the next section.
Part 4
The data achieved from the general formula will be compared with circles graphed using technology with different variables to find any potential scopes or limitations to the general formula.
By analysis the results achieved using the general formula and from the graphing technology, we can see that there are some limitations to the general formula. For OP = r -1 and r>OP, point A (intersection between circles C1 and C2) did not exist, hence there were no solutions to OP’. For these two limitations, the values achieved using the general formula did not match the values achieved using graphic technology.
However, there are several conditions which make the general statement, precisely when r = OP, OP > r, r = -OP and OP = -r as the results achieved from the general statement match the results achieved using the graphing technology.
From this data we can make a generalized statement that r cannot be greater that OP and OP cannot equal to the inverse of r. At some point, when r is greater than OP, there will be no intersection between the circles C1 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
