+704 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.
+1632 Let the unit circle's center be P, and the circle with radius 4, Q (the center).
AP = 1 because AP is the radius of the unit circle. BQ = 4 with the same logic. With a translation (mapping of points) to AB, P becomes point A, and Q is moved along the line BQ by 1. Let these points be P' = A, and Q'. PAB is a 90 degree angle because the radius is always perpendicular to the tangent, and QBA is 90 degrees. Thus P'BQ' is a 90 degree triangle with P'BQ' = 90 degrees, BQ' = 3, and P'Q' = 5. By pythagorean theorem, AB = 4.
+1632 Let the unit circle's center be P, and the circle with radius 4, Q (the center).
AP = 1 because AP is the radius of the unit circle. BQ = 4 with the same logic. With a translation (mapping of points) to AB, P becomes point A, and Q is moved along the line BQ by 1. Let these points be P' = A, and Q'. PAB is a 90 degree angle because the radius is always perpendicular to the tangent, and QBA is 90 degrees. Thus P'BQ' is a 90 degree triangle with P'BQ' = 90 degrees, BQ' = 3, and P'Q' = 5. By pythagorean theorem, AB = 4.
+129850 Thanks, proyaop...here's another way
(1) Extend BA in the direction of A and let it meet the extension of segment connecting the centers of the circle.....call this intersection point, O
Call the center of the larger circle M and the center of the smaller circle, N
Let x be the distance from the left edge of the small circle to O
Triangles BMO and ANO are similar such that
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 ] [ (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
