Aditya Gupta
Last Activity: 4 Years ago
dear student there is a theorem called Descartes' circle theorem for 4 kissing circles (you can google it and look up the wikipedia article for more info and refer the SPECIAL CASES section).
using that, we observe that (k1+k2+k3)^2= 2(k1^2+k2^2+k3^2) where k1, k2 and k3 are a, b, c resp
simplifying k1^2+k2^2+k3^2= 2(k1k2+k2k3+k3k1)
k1^2 – 2k1(k2+k3) + (k2 – k3)^2= 0
or k1= k2+k3 ± 2sqrt(k2k3)
k1= [sqrt(k2) ± sqrt(k3)]^2
or sqrt(k1)= sqrt(k2) ± sqrt(k3)
1/sqrt(a)= 1/sqrt(b) ± 1/sqrt(c)
since a＜b＜c, -ve sign is not possible
hence 1/√ a = 1/√ b + 1/√ c
KINDLY APPROVE :))
