Question 605925: I have a question that involves circles and chords and all the fun stuff. The question I have is actually a drawing so I'll do my best to explain it. There is a bigger circle that is beside a little circle and they touch at one point. The radius of the big circle is 5 and the radius of the little circle is 3. From a point at the top of the big circle to the top of the little circle there is a segment with the value of X. The question is asking to solve for X, Aka the segment. I know the answer is 2 radical 15 but I just need help figuring out how to show the work. I hope someone can help me THANKS!!
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Here's a drawing of what I'm seeing in my mind
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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Jim
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