Two circles with radii 5 and 3 are tangent externally. A common external tangent touches the circles at points A and B respectively. Find the length of AB.
Correct: C
Let the centers of the two circles be O1 and O2, and their radii be r1 = 5 and r2 = 3, respectively.
Since the circles are tangent externally, the distance between their centers is O1O2 = r1 + r2 = 5 + 3 = 8.
Let the common external tangent touch the first circle (center O1, radius r1) at point A and the second circle (center O2, radius r2) at point B.
The radii O1A and O2B are perpendicular to the tangent line AB.
Draw a line through O2 parallel to AB, intersecting O1A at point C.
This forms a rectangle O2BCA. Thus, AB = O2C and BC = O2A (no, BC is not O2A).
Instead, consider the right trapezoid O1ABO2 (where O1A and O2B are parallel). Draw a perpendicular from O2 to O1A. Let this point be C.
Then O1C = O1A - CA = O1A - O2B = r1 - r2 = 5 - 3 = 2.
And O2C is parallel to AB, so O2C = AB.
Now, consider the right-angled triangle O1CO2.
The hypotenuse is O1O2 = 8.
The leg O1C = r1 - r2 = 2.
The other leg is O2C, which is equal to the length AB we want to find.
By the Pythagorean theorem:
O1O2^2 = O1C^2 + O2C^2
8^2 = 2^2 + AB^2
64 = 4 + AB^2
AB^2 = 64 - 4
AB^2 = 60
AB = sqrt(60)
AB = sqrt(4 * 15)
AB = 2sqrt(15).