Two circles having radii #a#and #b# touch each other externally. If #c# is the radius of another circle which touches these two circles as well as a common tangent to the two circles, how do we prove that #1/sqrtc=1/sqrta+1/sqrtb#?

1 Answer
Jun 16, 2018

see explanation.

Explanation:

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Let #A,B,C# be the center of circle 1, circle 2 and circle 3, respectively, as shown in the figure.
#AB=a+b, AC=a+c, BC=b+c#
In #DeltaCDB, CD^2=BC^2-BD^2#
#=> CD^2=(b+c)^2-(b-c)^2#
#=> CD^2=(b+c+b-c)*(b+c-b+c)=2b*2c=4bc#
#=> CD=sqrt(4bc)=2sqrt(bc)#
Similarly, #CF^2=(a+c)^2-(a-c)^2#
#=> CF=2sqrt(ac)#
#=> DF=CF+CD=2sqrt(ac)+2sqrt(bc)=2sqrtc(sqrta+sqrtb)#
In #DeltaAEB, BE^2=AB^2-AE^2#
#=> BE^2=(a+b)^2-(b-a)^2#
#=> BE^2=(a+b+b-a)*(a+b-b+a)=2b*2a=4ab#
#=> BE=sqrt(4ab)=2sqrt(ab)#
As #BE=DF#,
#=> 2sqrt(ab)=2sqrtc(sqrta+sqrtb)#
#=> 1/sqrtc=(sqrta+sqrtb)/sqrt(ab)#
#=> 1/sqrtc=sqrta/sqrt(ab)+sqrtb/sqrt(ab)#
#=> 1/sqrtc=1/sqrtb+1/sqrta#
#=> 1/sqrtc=1/sqrta+1/sqrtb#