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Two circles are externally tangent at T.  The line AB is a common external tangent to the two circles, and P is the foot of the altitude from T to line AB. Find the length AB.

 

 Jun 23, 2024

Best Answer 

 #2
avatar+1230 
+1

We can set variables to complete this problem quite efiiciently. 

First, let's set the center of the large circle as M and the center of the smaller circle as N. 

Now, let's extend AB and and the line connecting the centers until they meet each other at a pont, O. 

Let the distance of the left edge of the smaller circle and O be x. 

 

Now, let's note that triangle BMO and ANO are similar. This is quite important. 

From this, we can write

\(BM / MO = AN / NO \\ 4 / (4 + 2 + x) = 1 / ( 1 + x) \\ 4 / ( 6 + x) = 1 /(1 +x) \\ 4(1 + x) = 1 (6 + x) \\ 4 + 4x = 6 + x \\ 3x = 2 \\ x = 2/3\)

 

Now that we have the value for x, we can easily solve for AB. We get that

\(OA = \sqrt{ NO^2 - NA^2 } = \sqrt{ (1 + 2/3)^2 - 1^2} = \sqrt{ 25/9 -1} = \sqrt{16/9} = 4/3 \\ OB = \sqrt{ MO^2 - MB^2} = \sqrt{ (6 + 2/3)^2 - 4^2 } = \sqrt{ 400/9 -16} = \sqrt{256/9} = 16/3 \\ AB = OB = OA = 16/3 - 4/3 = 12/3 = 4 \)

 

So our answer is 4

Feel free to let me know if I messed up!

 

Thanks! :)

 Jun 23, 2024
 #1
avatar+129725 
0

https://web2.0calc.com/questions/circles_146

 

 

cool cool cool

 Jun 23, 2024
 #2
avatar+1230 
+1
Best Answer

We can set variables to complete this problem quite efiiciently. 

First, let's set the center of the large circle as M and the center of the smaller circle as N. 

Now, let's extend AB and and the line connecting the centers until they meet each other at a pont, O. 

Let the distance of the left edge of the smaller circle and O be x. 

 

Now, let's note that triangle BMO and ANO are similar. This is quite important. 

From this, we can write

\(BM / MO = AN / NO \\ 4 / (4 + 2 + x) = 1 / ( 1 + x) \\ 4 / ( 6 + x) = 1 /(1 +x) \\ 4(1 + x) = 1 (6 + x) \\ 4 + 4x = 6 + x \\ 3x = 2 \\ x = 2/3\)

 

Now that we have the value for x, we can easily solve for AB. We get that

\(OA = \sqrt{ NO^2 - NA^2 } = \sqrt{ (1 + 2/3)^2 - 1^2} = \sqrt{ 25/9 -1} = \sqrt{16/9} = 4/3 \\ OB = \sqrt{ MO^2 - MB^2} = \sqrt{ (6 + 2/3)^2 - 4^2 } = \sqrt{ 400/9 -16} = \sqrt{256/9} = 16/3 \\ AB = OB = OA = 16/3 - 4/3 = 12/3 = 4 \)

 

So our answer is 4

Feel free to let me know if I messed up!

 

Thanks! :)

NotThatSmart Jun 23, 2024
 #3
avatar+129725 
0

Good job, NTS !!!

 

cool cool cool

CPhill  Jun 23, 2024
 #4
avatar+1230 
+1

 

Thank you! :)

 

~NTS

NotThatSmart  Jun 23, 2024
edited by NotThatSmart  Jun 23, 2024
edited by NotThatSmart  Jun 23, 2024

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