Maths Circles Ia
XYZ
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
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