# Ib Math Sl Ia - Circles

Topics: Circle, Analytic geometry, Triangle Pages: 7 (2436 words) Published: September 24, 2012
Math IA (SL TYPE1)
Circles

Circles
Introduction
The objective of this task is to explore the relationship between the positions of points within circles that intersect. The first figure illustrates circle C1 with radius r, centre O, and any point P. r is the distance between the centre O and any point (such as A) of circle C1.

Figure 1 The second diagram shows circle C2 with radius OP and centre P, as well as circle C3 with radius r and centre A. An intersection between C1 and C2 is marked by point A. The intersection of C3 with OP is marked by point P’. Figure 2

Through this investigation I will examine how the r values correlate with the values of OP in determining the length of OP’ when r is held as a constant variable and the value of OP is the variable that is subject to change. I will then venture on to study the inverse, the relationship when the r values becomes the variable that is changed and the OP value is held constant. r as a Constant

If we let the value of r be equal to 1, we can use that information to find the length of OP’ when OP=2, 3, and 4. The first thing one can deduce is that by using the points A, O, P’, and P two isosceles triangles can be formed; ∆AOP and ∆AOP’. To rationalize this assertion through an analytic approach it should first be understood that all line segments that connect the center of a circle to its perimeter, or circumference, is considered to be its radius. Because P’ and O are both located within the circumference of C3 and connect to its center, A, we can establish that AP’ and OA are radii of C3. Likewise, points O and A are situated within the circumference of C2 and connect to its center, P, meaning that both OP and AP can be regarded as the radius of C2. By definition all the radii within a same circle are of equal length, thus proving that OA and AP’ are of the same length, as well as AP and OP. This generates enough evidence to conclude ∆AOP and ∆AOP’ to both be isosceles triangles as is demonstrated in the diagrams below.

Knowing that there are two isosceles triangles, of which we know the length of two sides on each one, allows us a variety of methods we can employ to solve for the missing side using an algebraic process. At first, the sine/cosine rule, the Pythagorean Theorem, proportions, and basic trigonometry each seem like an appropriate approach to find the length of OP’, that is until we realize that all the circles are on a coordinate plane. Recognizing that all 3 circles are graphed leads to the comprehension that they can be given coordinates, the which can be plugged into the distance formula (d=(x2-x1)2+ (y1-y2)2) to find OP’.

Line OA is the radius of C3 and C1, which is set equal to 1. Because point O lies on the origin of the graph, its coordinates will be (0, 0). In this first scenario, Line AP, which is equal to OP and is the radius of C2, is going to be set at 2. Because point P is the ending point of a radii of C2 and lies on the x-axis, its coordinates will be (2, 0). We don’t know the coordinates of Point A so we will assign it variables, (a, b).

To solve for the coordinates of Point A we will first set the distance formula equal to 1 , the total length of OA, and plug in A(a, b) and O(0, 0). Once simplified, we should end up with 1= a2+b2. Then, so we can create a system of equations, we will solve for point A in line AP. This means we will set the distance formula equal to 2, the total length of AP, and plug in A(a, b) and P(2, 0). We should come up with 4= a2-4a+b2+4. Next, we subtract the first derived equation from the second to solve for the value of variable a in point A and find that a= 14. Through the method of substitution we are able to find that the value of variable b is ±154. Although the negative value is discarded since in the graph the value of variable b has to...

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