Let r=1, find OP’ when OP=2, OP=3, and OP=4.
Draw a perpendicular line,
∵
∴
∵
∴ When
=
∵
∵
∴
Link AP’, ∵p’ is on C3,
∴
, ∴
is an isosceles triangle.
∵
is an isosceles triangle,
∴
∴
Draw a perpendicular line,
∵
∴ When
=
∵
∵
∴
Link AP’, ∵p’ is on C3,
∴
, ∴
is an isosceles triangle.
∵
is an isosceles triangle,
∴
∴
Draw a perpendicular line,
∵
∴ When
=
∵
∵
∴
Link AP’, ∵p’ is on C3,
∴
, ∴
is an isosceles triangle.
∵
is an isosceles triangle,
∴
∴
Through the calculation, as we can see, when r=1,
We can notice that the length of OP’ gets small as the length of OP gets longer. Be specific, we can know that the length of OP’ is the reciprocal of the length of OP which is that OP’ is
, when OP is 2. From here, we can set a general statement, OP’=
. However, we have to realize that the length of r is 1, and this means the numerator of OP’ is not a just form of reciprocal number. Therefore, the general statement can be interpreted to another form, OP’=
.
Let’s consider another condition.
In the second condition, let OP=2, fixed, and find OP’ depending on the different lengths of r.
By the graph I’ve drawn, we can know that OP’ is on the same point of OP.
Therefore,
As the cosine rule,
Link
∵
∴
is an isosceles triangle, and
is an isosceles triangle
∵
∴
∴
∴
As the graph shown,
∵AO = 4 = r,
OP’ = 2AO = 2r
∴OP’ = 8
From the second assumption, we can find;
As the length of OP is stable and the length of r is changeable, the length of OP’ gets longer, which means the length of OP’ changes only by the changes in the length of r. In the first concept, the general statement I gave was OP’=
, however, it doesn’t match in this condition. If we follow the first general statement, the length of OP’ should be 1 when r=2. The same for the others, it should be
, and 2 when r=3, and r=4 respectively. Let’s make a diagram shown clearly.
From the diagram, we can figure out that the actual length of OP’ is calculated by multiplying one more time of r to the length of OP’ calculated which is followed the first general statement. Therefore, we can approach to the new general statement, which is OP’=
.
To test the validity of my general statement, I would like to change the values of OP and r. I will check up for the validity by using FOUR different conditions;
Before testing, I am going to estimate the results using the general statement. The lengths of OP’ are going to be 1,
,
and 9 respectively.
Let’s figure out the actual lengths of OP’!
Be Continued next page→

OP=4, r=2
Link AP’
is an isosceles triangle.
is an isosceles triangle.
As the Cosine Rule,
∵
∴
=
∴
= 4 + 4 – 8 (
)
= 8 – 7 = 1
The answer matches to my predicted value of OP’. Thus, for this condition, the general statement is valid.

OP=3, r=4
Link AP’
is an isosceles triangle.
is an isosceles triangle.
As the Cosine Rule,
∵
∴
=
∴
= 16 + 16 – 32 (
)
= 32 
=
In this condition, the estimated value and the actual value have the same value, so the general statement is, still, valid.

OP=5, r=7
Link AP’
is an isosceles triangle.
is an isosceles triangle.
As the Cosine Rule,
∵
∴
=
∴
= 49+49 – 98 (
)
= 98 –
=
The answer by calculation and the answer estimated was the same. This means the general statement is valid in this situation too.
Before we get into the 4th test, I will try to prove my general statement with unknown variables to show the validity.
I will assume that Circle 1(C1) has center O, Circle 2 (C2) has center P and Circle 3(C3) has center A. I will suppose that C1 has radius OP, and A is the intersecting point of C1 and C2. I will also presuppose that C3 has its radius, r, and the point P’ is the intersection of C3 with OP. Note that the figure is shown three intersecting circles with different unknown variables. The graph is drawn randomly, not scaled.
Link AP’
is an isosceles triangle.
is an isosceles triangle.
It is not possible to draw three circles required.
Through the 4 different types of test, we can notice that the general statement is valid on first three tests, but not for the last one. The only characteristic I found out was that the length of OP is shorter 2 units than the length of r. But, I also realized that the third condition has the same lengths of OP and r the last condition does have. I have tried to figure out how C1 and C2 couldn’t intersect at one point. First of all, I organized the lengths of OP and r by graphing a diagram:
According to the diagram shown, we can know that C1 and C2 don’t intersect at one point when OP=
r. This tells us about the length of OP should be longer than
r. I assume one more situation which has the length of OP is 1 and the length of r is 2. In this situation, OP=
r. I tried to draw the circles, and it was possible to draw three circles.
Now, what I have is that the length OP should be longer than
r, but it’s possible to equal
r. Then, let K denote the coefficient of r, set OP=Kr and test what value of K would be, whether three circles can be drawn. I would like to find the value of K which is greater than
and smaller than
,
. I assume a situation which is OP of 5, r of 12.
The graph shows, if K=
, so OP=
r, then C1 and C2 couldn’t intersect at one point, therefore, it is impossible to draw three circles required.
Since OP and r are the radii of circles, so their values can’t be negative. Moreover, the length of OP should not be less than
r, the limitation of the general statement is
r.
As conclusion, I could find out the general statement by two different conditions which have different fixed value of OP and r, and testing the validity of the general statement I found. Thus, the general statement I found finally is OP’=
. However, the statement also has a scope, which is
r.