Question

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`?

Answer 1

see explanation.

Explanation:

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`