As the figure above shows, two circles lie on a line and tangent to ea

Wonderwoman31 , let me know if the approach below does not make sense.

stne , it's hard to tell exactly, but the question tests tangent lines, so A and B are the points of tangency on the exterior tangent line.

MBA20 , as requested, no formulas. (I don't blame you. I don't use distance formulas when I can avoid doing so. This distance formula is very handy. It's in the footnote*, but as you can see, you don't need it.)

Essentially, connect the centers; make a quadrilateral using perpendicular radii; divide that quadrilateral into a rectangle and a right triangle; and solve for the unknown side length of the right triangle (MQ). That length = AB.

Properties of two circles that are tangent to each other
• The circles touch each other at exactly one point and lie on the same line PQ
• The circles have three common tangent lines: 1) one internal "common" tangent line and 2) two external tangent lines (see little drawing in blue, lower left of diagram).
For these two circles, the bottom line (LINE AB) is an exterior tangent line.
• The radius from the center of each circle is perpendicular to the tangent line, on which lies the point of tangency.
• Derive parallel and perpendicular lines in the diagram from the statement directly above

Circles P and Q are both tangent to LINE AB at points A and B, respectively.

Make a quadrilateral, then make a right triangle.
Divide the quadrilateral into a rectangle and a right triangle.

1) Connect P and Q

2) Drop a line from P to A and from Q to B
-- Those lines are perpendicular to LINE AB.
PA and QB are radii of each circle and therefore perpendicular to LINE AB at points A and B
-- Those radii are parallel: each radius PA and QB is perpendicular to LINE AB. Two lines that are perpendicular to the same line are parallel to each other.

3) Now we have a quadrilateral APQB

4) Draw a line from the center of Q to point M; that line is parallel to the bottom line (external tangent line AB)

5) We have rectangle AMQB and right ∆ MPQ

6) The base of the right triangle, MQ, is the length of AB (rectangle has sides of equal length)

7) Let AB = MQ = x

Right ∆ MPQ

• Length of side PQ, hypotenuse \(= (2r + r) = 3r\)
• Length of side MP \(= (2r - r) = r\)
-- Subtract the rectangle side length from the radius to get MP, that is
-- (PA - MA)= MP
PA = \(2r\)
MA = QB = \(r\)

• use Pythagorean theorem to find length of third side, x

Pythagorean theorem

\(MP^2 + x^2 = PQ^2\)
\(r^2 + x^2 = (3r)^2\)
\(r^2 + x^2 = 9r^2\)
\(x^2 = 8r^2\)
\(\sqrt{x^2} =\sqrt{8r^2}\)
\(x = 2r\sqrt{2}\)

Rewrite:
\(x\) = MQ = AB = \(2\sqrt{2}*r\)

ANSWER C

Hope that helps.


*From what I just did, you can derive a simpler version of the formula that m1033512 used +1.
\(a\) = longer radius (in this problem, \(a = 2r\))
\(b\) = shorter radius (in this problem, \(b = r\))

The distance between base points of two circles tangent to each other is
\(\sqrt{(a+b)^2 - (a-b)^2}=2\sqrt{ab}\)

EDIT: nick1816 , +1.