SOLUTION: draw a big circle 1 and draw another smaller circle 2 to the right of the big circle 1 and touches the big circle 1. Draw a line that tangents to both circles at the bottom. Draw a

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Question 55924: draw a big circle 1 and draw another smaller circle 2 to the right of the big circle 1 and touches the big circle 1. Draw a line that tangents to both circles at the bottom. Draw a third smaller circle 3 in the space between the two larger circles and the line so that the third circle is touching the two bigger circles and tangent to the line. r1, r2, and r3 are radii of circle 1, circle 2, and circle 3. Express r3 in terms of r1 and r2.
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
draw a big circle 1 and draw another smaller circle 2
to the right of the big circle 1 and touches the big 
circle 1. Draw a line that tangents to both circles 
at the bottom. Draw a third smaller circle 3 in the 
space between the two larger circles and the line so 
that the third circle is touching the two bigger 
circles and tangent to the line. r1, r2, and r3 are 
radii of circle 1, circle 2, and circle 3. Express 
r3 in terms of r1 and r2.

Let A be the center of the circle 1, B the center of 
circle 2, and C the center of circle 3.

Draw triangle ABC.  

Draw the radii AD, CE and BF perpendicular to the 
tangent line, of which DE and EF are segments.

Draw a line thru C parallel to the tangent line.  
Let it intersect AD at G and BF at H.

AG = r1-r3, BH = r2-r3

AC = r1+r3, BC = r2+r3

Triangles AGC and BHC are right triangles, so we 
can use the Pythagorean theorem:
      _______    _________________    _____     ____
GC = ÖAC²-AG² = Ö(r1+r3)²-(r1-r3)² = Ö4r1r3 = 2Ör1r3       
      _______    _________________    _____     ____
HC = ÖBC²-BH² = Ö(r2+r3)²-(r2-r3)² = Ö4r2r3 = 2Ör2r3
                   ____     ____     __  __    __
Now GH = GC+HC = 2Ör1r3 + 2Ör2r3 = 2Ör3(Ör1 + Ör2)

Draw a line thru B parallel to GH and DF. Let it 
intersect AD at I. 

AB = r1+r2,  AI = r1-r2

Triangle AIB is a right triangle,  so we can use 
the Pythagorean theorem:
      _______    _________________    _____     ____  
IB = ÖAB²-AI² = Ö(r1+r2)²-(r1-r2)² = Ö4r1r2 = 2Ör1r2

Now IB = GH so
  ____     __  __    __   
2Ör1r2 = 2Ör3(Ör1 + Ör2)

Divide through by 2
  ____    __  __    __   
 Ör1r2 = Ör3(Ör1 + Ör2)
 
Square both sides:
           __    __  
r1r2 = r3(Ör1 + Ör2)²

        r1r2    
r3 = ------------
     (Ör1 + Ör2)²

Edwin


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