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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.

 May 29, 2024
 #1
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First, we can graph this problem. 

 

We can use the Pythagorean Theorem to complete this problem. 

First, we need similar triangles. We can extend AB towards the direction of A and meet it up with the line connecting the radiuses. We can call this point O. 

Let the center of the unit circle be N and the center of the larger circle M. Let's let x be the distance from the left edge of the smaller circle to O. 

 

Notice that triangles BMO and ANO are similar triangles. This means we can write

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

 

You can also use translation for this problem, but my small brain can't figure it out!

Thanks! :)

 May 30, 2024

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