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MATH IA: investigate the position of points in intersecting circles

Extracts from this document...

Introduction

Julio Francati

Investigate the position of points in intersecting circles

IB analytical Geometry

Aim:

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

Introduction:

The following diagram shows a circle C1 with center O and radius r, and any point P.

image38.png

Circle C2 has center P and radius OP. Let A be one of the points of intersection of C1and C2. Circle C3has center A, and radius r. The point P’ is the intersection of C3 with (OP). This is shown in the diagram below

image39.png

In order to address the aim of this Internal Assessment, the different situations of C1, C2, C3 need to be discussed and investigated. The knowledge used in this investigation is the Pythagorean Theorem. This theorem states that in a right triangle the hypotenuse is equal to the sum of the legs of the triangle squares. The equation used for this theorem is a2+b2=c2. The next form of knowledge used is a modified version of the cos θ rule. The formula for the

...read more.

Middle

GP=4-x.

image03.png

OAG and ΔAGP are both right triangles that share the side AG. To solve for x we will use the Pythagorean triangle and set the two legs of each triangle equal to each other. This can be seen as:

image13.png

image14.png

              Get rid of the radicals.

image15.png

                                Add x2 to both sides

image16.png

                                   Solve for x

OP’=2GP= image17.png

After OP’ is found I’m going to attempt to find a pattern so a general statement can be found.

OP

2

3

4

OP’

image02.png

image12.png

image17.png

r

1

1

1

After looking at the results it can be seen that the general statement is:image18.png

.

In the second part of this investigation r is now the variable and OP will have a controlled value at 2.

When r=2

image43.png

After making an accurate graph point P and P’ lie on the same point. As a result

OP=OP’=2.

When r=3

image44.png

After making the graph a line is drawn to link AP’ and AP. Since OA and AP’ are in the same circle they are equal to each other making image01.png

AOP=image01.png

AP’O.

OA=3

OP=2=AP

Because PO and PA are in the same circleimage01.png

AOP=image19.png

. As a result OP=OA, image01.png

POA=image01.png

OAPimage01.png

41.41image03.png

image01.png

PAP’=180image20.png

41.41image21.png

Thus, PP’image22.png

OP=2+PP’=2+2.51=4.51

...read more.

Conclusion

image23.png

) the value matches the value of OP’ making the general statement valid for this set of values.

Test 2:

When r=3,OP=4

image46.png

First connect AP’

In image03.png

AOP’ and image03.png

PAO

OA=AP’=3

AP=OP=4

image01.png

AOP’=image01.png

AP’O   image01.png

PAO=image01.png

AOP’

Therefore image01.png

AOP’=image01.png

AP’O=image01.png

PAO

In image03.png

AOP’ and image03.png

PAO, two angles are equal so the last angle must be equal as well.

Therefore,

image01.png

AOP’=image01.png

PAO

image27.png

AP’O=image01.png

AOP’

image01.png

OAP’=image01.png

OPA

Consequently, image03.png

AOP’ and image03.png

PAO are similar triangles.

The Similarity ratio of the two triangles is image30.png

OP’= image31.png

 and the given values are r=3,OP=4. When plugged into the general statement (image23.png

) the value matches the value of OP’ making the general statement valid for this set of values.

In conclusion this general statement has both limitations and scope. One limitation that can be seen is that for every trial OPimage32.png

. Another factor that must be taken into consideration is that OPimage33.png

, this is the only way that allows C2 and C3 to have at least one intersection. Therefore the overall scope of this investigation is OPimage33.png

. The formation of the general statement is seen below.

image03.png

AOP’ and image03.png

PAO are similar triangles. Therefore: image34.png

Cross multiply: OP’image35.png

OP’= image36.png

OA=r, AP=OP

After making the proper replacements and implementing the scope the general statement is

image23.png

, image37.png

.

...read more.

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