# Math Portfolio Ib Circles

MATHS PORTFORLIO SL TYPE I CIRCLES ? In this portfolio I am investigating the positions of points in intersecting circles. (These are shown on the following page. The following diagram shows a circle C1 with centre O and radius r, and any point P. The circle C2 has centre P and radius OP. Let A be one of the points of intersection of C1 and C2. Circle C3 has the centre A, and radius r. The point P? is the intersection of C3 with (OP). This is shown in the diagram below. As shown on the assignment sheet, r=OA. We therefore need to find the values of OP? hen r=OA=1 for the following of the values of OP: OP=2, OP=3 and OP=4. We first of all extract the triangle OPA from the above diagram and since we have the values for all the three sides we can finds the angle AOP which will later on help to get the value of length OP?. The circle C2 and triangle OPA are shown below with all side of OPA indicated. OP=AP since they are the radii of the same circle, C2. Having all the three sides, we can now calubulate the angle AOP using the cosine rule. Angle AOP is calculated below: Cos AOP=(2^(2 _ ) 2^(2 _ ) 1^2)/(-2? 2? 1) Cos AOP=0. 25 ? AOP =COS-10. 5 =75. 52248781 ?75. 5? Since we now having the triangle AOP, we can extract the triangle AOP? from the diagram shown on the previous page which in return will help us to find OP? using the sine rule. The triangle AOP? is shown below: O P For accuracy the value of angle AOP will be used as cos-10. 25 instead of 75. 5?. Since triangle AOP? is an isosceles triangle, AOP=AP? O=cos-10. 25. Therefore OAP? = (180-(2? cos-10. 25)). The calculation of the value of OP? is shown below: (OP? )/(sinOAP? )=(AP? )/(sinAOP? ) (OP? )/sin? (180-(2?? cos? ^(-1) 0. 25) ) =1/sin? (? cos? ^(-1) 0. 5) op? = 1/sin? (0. 25) ? (180-(2?? cos? ^(-1) 0. 25) ) op? =1/2 When OP=3, the triangle OPA and the calculation of OP? are as follows: Cos AOP = (3^2-3^2-1^2)/(-2? 3? 1) Cos AOP = 1/6 ? AOP = cos-1 1/6 =84. 4? From the triangle AOP? we can now calculate the length of OP? using the sine rule as before. The triangle AOP? and the calculator of OP? is shown below: A 16 OP’ AP? O=AOP? =cos-11/6 OAP? = (180-(2?? cos? ^(-1) 1/6) (OP? )/(sinOAP ? )= (AP? )/(sinAOP? ) (OP? )/sin? (180-(2?? cos? ^(-1) 1/6) ) =1/sin? (? cos? ^(-1) 1/6) op? = 1/sin? (1/6) ? (180-(2?? cos? (-1) 1/6) ) = 1/3 When OP=4; Cos AOP =(4^2-4^2-1^2)/(-2? 4? 1) Cos AOP = 1/8 ? AOP =cos-1 1/8 =82. 4? Using the sine rule; (OP? )/(sinOAP ? )= (AP? )/(sinAOP? ) (OP? )/sin? (180-(2?? cos? ^(-1) 1/8) ) =1/sin? (? cos? ^(-1) 1/8) op? = 1/(sin? (? cos? ^(-1 ) 1/8))? (180-(2?? cos? ^(-1) 1/8) ) = 1/4 When OP=2, OP? = 1/2;when OP=3 , OP? = 1/3 and when OP=4, OP? = 1/4 . This indicates that the value of OP? is dependent on the value of OP. In fact it is inversely proportional to the value of OP.

To arrive at the value of OP? , 1 is divided by the value of OP. Therefore generally, the value of OP’ can be written as: OP=r/OP Moreover, from the values of OP? calculated above, it is observed that the value of OP? is twice Cos AOP. The general statement therefore can be written as: OP? = 2 Cos? Let OP=2. Find OP? when r=2, r=3 and r=4. Describe what you notice and write a general statement to represent this. Comment on whether or not this statement is consistent with your earlier statement. First of all we need to calculate the value of OP? when OP=2 and r=2. The triangle AOP now looks like as follows:

A 22 1. 5P’ Since all the sides are of the same length, then AOP=APO=OAP=60° (according to the angle sum of the triangle). The triangle AOP? is shown below from which OP? is found. A 22 O2P’ AO=AP? from the diagram given on the lab sheet, therefore AOP=APO=60°. The remaining angle OAP= (180-(2? 60)) = 60°. This then means that triangle AOP? is an equilateral triangle – all its sides are the same. ?AO=AP? =OP? =2 We now need to calculate the value of OP? when OP=2 and r=3. Below is the triangle AOP and the calculation of angle AOP. 22 3 Cos AOP = (2^2-2^2-3^2)/(-2? 2? 3) Cos AOP = 3/4 AOP = cos-1 3/4 = 41. 4° Having calculated the value of angle AOP, we can now calculate the value of OP? from AOP? using the sine rule as shown below: A 33 O P? Angle OAP? = 180-(2?? Cos? ^(-1) 3/4) (OP? )/(sinOAP ? )= (AP? )/(sinAOP? ) (OP? )/sin? (180-(2?? cos? ^(-1) 3/4) ) =1/sin? (? cos? ^(-1) 3/4) op? = 1/sin? (? cos? ^(-1) 3/4) ? (180-(2?? cos? ^(-1) 1/8) ) =9/2=4. 5 The value of OP? is now calculated when OP=2 and r=4 using the same method as above. 22 O4P 44 P? Cos AOP = (2^2-2^2-4^2)/(-2? 2? 4) Cos AOP = 1 ? AOP = cos-1 1= 0

AOP? =AP? O=0 ?OAP? =180-(2? 0) = 180° OP? 2= OA2+ AP? 2-2? OA? AP? Cos OAP? = 42+42-2? 4? 4 Cos 180° = 64 OP? = v64 = 8 Below is the table for the vales of r, OP and OP’ when OP is kept constant. ROPOP’ 222 324. 5 428 When OP=2, OP? =2; when OP=3, OP? = ( 9)/2 and when OP=4, OP? =8. From these results it can be seen that the length OP? increases with the increasing length OP and the general statement for the variation in the values of OP? is as follows: OP? = r^2/OP. This general statement is not fully consistent with the first one because r/OP is not always equal to r^2/OP.

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When the above values are substituted into the first general statement, wrong values of OP? are obtained but the latter general statement holds for both data. However both the general statements hold true when r=1 since 12=1, which indicates that for this value of r, r^2/n=r/n. Use technology to investigate other values of r and OP. Find the general statement of OP?. I used GeoGebra to draw the intersecting circles with the values of r and OP stated and the values of OP? were automatically calculated. When r=1and OP=2 OP=0. 5

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