Two circles having radii aaand bb touch each other externally. If cc 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/sqrtb1c=1a+1b?

1 Answer
Jun 16, 2018

see explanation.

Explanation:

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Let A,B,CA,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+cAB=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