Question 605925
Here's a drawing of what I'm seeing in my mind

<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/Algebra%20dot%20com/tangent_circles4.png">

In the drawing, I added the point F to be the point of intersection between the common tangent line of the two circles and the line that passes through the two centers.
I've added the additional line segments DF (length y) and BF (length z). 
I've also labeled every important piece of information.
In addition, I've made notes as to what the lengths of each segment are (whether known or unknown)



The first thing to do is to find the length of z. This will help us find the value of line segment AF (which is the hypotenuse of the larger right triangle ACF)



It helps to remember that ALL circles are similar (since they have the same basic shape). 
So this means that the radii in the larger circle are all proportional to the radii in the smaller circle (or vice versa)


This translates to the fact that triangle AFC is similar to triangle DFB. 
This is because the two triangles share a common angle (at point F) and the segments are proportional.


So because the two triangles are similar, we can say


AF/BF = AC/BD


But we know from the drawing that AC = 5 and BD = 3. 
Furthermore, we let BF = z.
 
The length of AF is simply the sum of the segments AE, BE, and BF, so 


AF = AE + BE + BF

AF = 5+3+z

AF = 8+z


This means that we can then plug all of this information into AF/BF = AC/BD



AF/BF = AC/BD


(8+z)/z = 5/3


From here, solve for z


(8+z)/z = 5/3


3(8+z) = 5z


24+3z = 5z


24 = 5z-3z


24 = 2z


24/2 = z


12 = z


z = 12


So the length of z is z = 12. This means segment BF is 12 units long (ie BF = 12)


So the length of segment AF is 


AF = 8+z


AF = 8+12


AF = 20


These two pieces of info (that BF = 12 and AF = 20) will be very important in finding x and y.


Let's use the fact that BF = 12 to find the value of y.


From the pythagorean theorem, we can say


BD^2 + DF^2 = BF^2


plug in BD = 3, DF = y and BF = 12 and solve for y


3^2 + y^2 = 12^2


9 + y^2 = 144


y^2 = 144 - 9


y^2 = 135


y = sqrt(135)


y = 3*sqrt(15)


So DF = 3*sqrt(15) units


Now use the facts that AF = 20, AC = 5 and CF = x+3*sqrt(15) to get



AC^2 + CF^2 = AF^2


5^2 + (x+3*sqrt(15))^2 = 20^2


25 + (x+3*sqrt(15))^2 = 400


(x+3*sqrt(15))^2 = 400-25


(x+3*sqrt(15))^2 = 375


x+3*sqrt(15) = sqrt(375)


x+3*sqrt(15) = 5*sqrt(15)


x = 5*sqrt(15) - 3*sqrt(15)


x = (5 - 3)*sqrt(15)


x = 2*sqrt(15)


So the exact length of segment CD is 2*sqrt(15) units.


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