In the figure above, line segment \(AD\) is the diameter of circle \(O,\) line segment \(AO\) is the diameter of circle

[Gmat_mission]

In the figure above, line segment \(AD\) is the diameter of circle \(O,\) line segment \(AO\) is the diameter of circle \(B,\) line segment \(OD\) is the diameter of circle \(C,\) and circle \(E\) is tangent to each of the other circles. If the radius of circle \(O\) is \(4,\) what is the radius of circle \(E?\)

A. \(2/3\)

B. \(3/4\)

C. \(1\)

D. \(4/3\)

E. \(3/2\)

[spoiler]OA=D[/spoiler]

Source: Veritas Prep

[Scott@TargetTestPrep]

FourCircles.png

In the figure above, line segment \(AD\) is the diameter of circle \(O,\) line segment \(AO\) is the diameter of circle \(B,\) line segment \(OD\) is the diameter of circle \(C,\) and circle \(E\) is tangent to each of the other circles. If the radius of circle \(O\) is \(4,\) what is the radius of circle \(E?\)

A. \(2/3\)

B. \(3/4\)

C. \(1\)

D. \(4/3\)

E. \(3/2\)

[spoiler]OA=D[/spoiler]

Solution:

Since the radius of circle O is 4 (i.e., AO = 4), BO = 2. We can form a right triangle by connecting B, E, and O. If we let x = radius of circle E, the legs of right triangle BOE are BO = 2 and OE = 4 - x, and its hypotenuse is BE = 2 + x. Now, using the Pythagorean theorem, we have:

2^2 + (4 - x)^2 = (2 + x)^2

4 + 16 - 8x + x^2 = 4 + 4x + x^2

16 - 8x = 4x

16 = 12x

x = 16/12 = 4/3

Answer: D



Source: Veritas Prep