Geometry Problem Involving Circles and Fixed Radius-

[Canada_Whiz]

Geometry Problem Involving Circles and Fixed Radius- Please Help!

I have a compass with radius X, and my friend has a compass with radius Y. We both draw a circle with our compasses so that our circles intersect at two points. Call these two points C and D.
You draw a common tangent to both circles, meeting my circle at point A and my friends circle at point B. Prove that the circles passing through A, B and C are the same, regardless of where my friend and I draw our circles, and they are the same as the radius of the circle going through A, B and D.




I can solve this problem for X=Y, but i don't know how to do any other cases

 

[zgozvrm]

Prove that the circles passing through A, B and C are the same

I don't quite understand the question. There is only one circle that will pass through any 3 given points. So, what are we to prove that it is the same as?

 

[Canada_Whiz]

You have to prove the radius is the same regardless of how my friend and I draw our circles. So like if we put the circles really close or really far apart, it will be the same radius (given that the circles still intersect at two points)

 

[Quinzio]

Ok, so write down the coordinates af A, B, C, D given the radii X, Y and distance from centers W.

 

[Canada_Whiz]

Ok i did that and got INCREDIBLY messy coordinates for the intersection points, not to mention I have no idea how to even find the tangent points of the common tangent.

There has to be a purely synthetic solution. Thanks :)

 

[Quinzio]

Try putting the center of one circle in the origin an the center of the other on the x axis.
This should simplify a lot.

What are anyway the messy equations.
I don't think you should get so messy stuff.

 

[zgozvrm]

We need to show that radius R (of the circle through points A, B & C) is purely a function of X and Y, independent of W (the distance between them).

I've come up with an equation (actually, a spreadsheet) for R, based on X, Y & W, that will produce the same value of R for any W entered into the equation,
such that X<Y & (X - Y) < W < (X + Y). But it isn't a proof. I need to eliminate the W factor from the equation...

 

[Quinzio]

We need to show that radius R (of the circle through points A, B & C) is purely a function of X and Y, independent of W (the distance between them).
.

No, the radious R is indeed dependent from W, how else can it be.

You need to show that: the circle passing through A,B,C has the same radious than the circle passing through A,B,D.

As I show in the picture.
Circle in thick line are the original circle. A tangent is drawn. The circles in orange and blue have the same radious, as you need to prove.

[PLAIN]http://img836.imageshack.us/img836/5773/provaed.jpg

 

[Quinzio]

Different original circles.

The blue and red circles have still the same radious.

[PLAIN]http://img89.imageshack.us/img89/8733/provaa.jpg

 

[zgozvrm]

No, the radious R is indeed dependent from W, how else can it be.

You need to show that: the circle passing through A,B,C has the same radious than the circle passing through A,B,D.

First of all, there are 2 things that need to be shown:
1) that the radius of the circle passing through the point A, B, and C remains constant regardless of the distance W between the centers of the circles
2) that the radius of the circle passing through points A, B, and C is equal to the radius of the circle passing through points A, B, and D

Therefore, radius R CANNOT BE dependent on distance W. If it where then the distance would have an effect on the length of R which is what we are proving is not the case!

 

[Quinzio]

1) that the radius of the circle passing through the point A, B, and C remains constant regardless of the distance W between the centers of the circles

This is not true.
If the radious of the 2 original circles do not change and only W change, this is not true.
It's really evident that is not true.

Consider when RA + RB < W, the two circles don't intersect.
So how can be said the W is not relevant ?

 

[zgozvrm]

Quinzio, I don't know how you got your points where they are, but this is how everything should be labeled (note that the distance W is from point X to point Y, the centers of the cirlces):

 

[zgozvrm]

Re-read the question, specifically:

Prove that the circles passing through A, B and C are the same, regardless of where my friend and I draw our circles, and they are the same as the radius of the circle going through A, B and D.

This is not true.
If the radious of the 2 original circles do not change and only W change, this is not true.
It's really evident that is not true.

It is true.
We are being asked to prove that the radius of the circle passing through points A, B, and C will have a constant radius, regardless of how far apart the circles are (as long as they intersect in 2 places)

 

[Quinzio]

Ok, so if it's true, you just need to prove it. :)

 

[Quinzio]

Quinzio, I don't know how you got your points where they are, but this is how everything should be labeled (note that the distance W is from point X to point Y, the centers of the cirlces):

Ok, in my drawings the name of the points are not the ones used in the discussion.

 

[zgozvrm]

Ok, so if it's true, you just need to prove it. :)

It IS true (you can use AutoCAD to verify it). And proving it is the whole point of this post!
Being a homework forum, we're only supposed to help the OP, not give them the answer.

 

[Quinzio]

Let's make 2 examples.
Example A
Circle 1 center (0,0) radious 4
Circle 2 center (5,0) radious 4
Centers distance 3

Example B
Circle 1 center (0,0) radious 4
Circle 2 center (7,0) radious 4
Centers distance 7This 2 examples differ only by the distance of the centers.

Are you saying that the resulting circles passing through A, B and C have the same radious in the 2 examples ?

 

[zgozvrm]

Let's make 2 examples.
Example A
Circle 1 center (0,0) radious 4
Circle 2 center (5,0) radious 4
Centers distance 3

Example B
Circle 1 center (0,0) radious 4
Circle 2 center (7,0) radious 4
Centers distance 7


This 2 examples differ only by the distance of the centers.

Are you saying that the resulting circles passing through A, B and C have the same radious in the 2 examples ?

Yes, that's exactly what I'm saying (try it yourself!). And this is what the OP is trying to prove.
But he's already done the proof for the case where X = Y, which is your example (Circle 1 has radius X = 4, Circle 2 has radius Y = 4).
Now, he needs to prove the case where X \ne[/tex] Y, or more simply, for X &gt; Y

 

[zgozvrm]

In other words, if you have:

Case A
Circle X with center (0, 0) and radius x = 4
Circle Y with center (5, 0) and radius y = 3
(centers have distance of w = 5)

Case B
Circle X with center (0, 0) and radius x = 4
Circle Y with center (6, 0) and radius y = 3
(centers have distance of w = 6)

The circle that passes through points A, B & C in example A will have the same radius as the circle that passes through points A, B & C in example B.

We need to prove this for all cases where circle 1 and circle 2 intersect in 2 points, which is when (x - y) < w < (x + y) --- in this example 1 < w < 7

 

[Canada_Whiz]

Thanks for all the interest in this problem, but does anyone know where to start? Coordinate bashing is WAY too ugly.

 

[Canada_Whiz]

To prove the circles formed by ABC and ABD are the same, we just need to prove that <ACB+<ADB=180. Can anyone tell me how to do that?

 

[zgozvrm]

Try constructing radius lines from the center of the circle passing through A, B & D to points A & B, then use the inscribed angle theorem

 

[Quinzio]

It appears that the radious of circle passing through ABD / ABC has a radious sqrt(X*Y)

Incredible.

If the proble is just to show that ( radious ABD = radious ABC ) I think it's obvious because the drawing is very simmetrical.

But now I want to show that radious ABC = sqrt(X*Y)
But how ?