# Circles Portfolio. The purpose of this assignment is to investigate several positions of points in three intersecting circles in order to discover a general statement

Mathematics SL, Portfolio Type 1Circles

CIRCLES

The purpose of this assignment is to investigate several positions of points in three intersecting circles in order to discover a general statement. There are three circles: circle C1 has center O and radius r, circle C2 has center P and radius OP, and circle C3 has center A and radius r (the same radio as C3). Point P' is the intersection of C3 with (OP).

All diagrams in this portfolio were created with Geogebra. To give a clear display of the length of each segment the diagrams are presented in the Cartesian coordinate system using point O as origin and (OP) as x-axis.

At first I will try to find a general statement representing the relation of OP’, OP and r in three examples, in which r will remain constant and OP will have three different values.

Diagram  1: r=1, OP=2

The method I will use to find OP’ will include the cosine function used in two triangles I will sketch in the first three diagrams:

and

.

In diagram 1:

Triangle OPA                           Triangle OP’A                                                                                                 a₁=OP=2                                   a₂=OP’=?                                                                                                                                                 o= PA=2                                   o₁=P’A=1                                                                                                                                                                                     p=p’=AO=1                              p’=p=AO=1

To find OP’ with the cosine function we must know the angle measurement of OAP’ (the opposite angle). I will find OAP’ through different steps involving both triangles. I will begin with finding OAP:

a₁²=o²+p²-2·o·p·

A=cosˉ¹ (

)

A= cosˉ¹ (

)

A=75.52ᵒ

Because

is isosceles, OAP=AOP=75.52ᵒ.

AOP is shared by both triangles, which means that OP’A is 75.52ᵒ as well. From this we can assume that OAP’=180ᵒ-2·75.52ᵒ=28.96ᵒ=A₂

Now I can easily find the length of OP’:

a₂²= (OP’) ²=

)

a₂=OP’=

a₂= OP’=

a₂= OP’=0.50

In diagram 2:

Triangle OPA                           Triangle OP’A                                          ...