IB SL Math

Circles

Aim:

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

The above diagram shows that distance r is the distance between any point, such as A, and the center of the circle, O, of the circle C1. The circle C2 has centre P and radius OP. A is one of the points of intersection of C1 and C2. Circle C3 has centre A, and radius r. The point P’ is the intersection of C3 with (OP). The r=1, OP=2, and P’=0.5. This is shown in the diagram below.

This experiment will explore the relationship between the OP values and the r values, when the r values are held constant. It will also investigate the reverse, the relationship when the OP values are held constant and the r values are modified. In the first example, the r value is 1 unit. An analytic approach will be taken to find the length of OP’ when OP=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 OA and AP’ have the same lengths because both points O and P’ are within the circumference of the circle C3, which means that OA and AP’ are its radius. Similiarly, ΔAOP forms another iscosceles triangle where AP and OP are equal in length, because both OP and AP are within the circumference of the circle C2. OA = r of C3 or C1= 1. Since the circles have been graphed, their points can be denoted as coordinates. For OA, the coordinates of O will be (0,0), because it lies in the origin of the graph. The coordinates of the point are undefined, so they shall be denoted as (a,b). We need these coordinates because to find the value of P’. AP = OP = r of C2 = 2. From this, we can get the coordinated of AP. The coordinated for the point P will be (2,0), for it lies on the x-axis and has a radius of 2. A is still undefined, so we will leave it as (a,b). Using the distance formula, (a-c)2+ (b-d)2, we can use an algebraic method of solving equations to solve for the unknown coordinates. Using this, we can find the coordinates for A. OA = 1, A = (a,b), O = (0,0)

By the distance formula:

1= (a-0)2+(b-0)2

1= a2+b2

(1=a2+b2)2

1= a2+ b2

and,

AP= 2, A(a, b), P(2, 0)

Then, by the distance formula:

2= (a-2)2+(b-0)2

(2= √(a2+ 4a+4+ b2 ))2

4= a2+ 4a+4+ b2

Using the elimination method, the value of a can be found.

1= a2+ b2

- 4= a2+ 4a+4+ b2

_________________________

1= 4a

a= 14

substituting this value of a in previous equation for finding the value of b: 1= a2+ b2

1= 142 + b2

b2 = 1- 116

= 1516

b = ±1516 = ±154

We will not take the value of negative b, however, because in the graph, b falls in quadrant one, where the y value is positive. So, the coordinates for point A are, A = ( 14, 154)

Now, we have to find the coordinates of point P’. for this, we will use the distance formula and take P’= (c,0). We know the length of AP’ because it is the radius of C3, which is 1. Since we know the coordinates of A, we sub in the numbers in the distance formula. AP’= 1 = c- 142+0-1542

= c- 142+1542

1=c- 142+1542 ]2

1= (c-14)2 + (154)2

1－ (154)2 = (c-14)2

c-14 = 116

c= 14 ± 14

c= 12 , 0

According to the graph, the point P’ does not lie on the origin, so the 0 can be neglected. This means that the coordinates of the point P’ are ( 12 , 0). Now, the length of OP’ can be calculated by using the distance formula for the two points A and P’. So, (0-12)2+ (0-0)2

= (12)2= 12

∴ OP = 2, r = 1, OP'= 12

Using the same method of calculation for the other values, the following is obtained: r| OP| OP’|

1| 2| 12|

1| 3| 13|

1| 4| 14|

The relationship between OP and OP’, when r is held constant, is inversely proportional. So when the value of one of them increases, the value of the other decreases proportionally. Therefore, the general...