Three circles of radii a, b, c (a＜b＜c) touch each other externally. If they have x axis as their common yangent, then: 1/ √ a = 1/ √ b + 1/ √ c 1/ √ b = 1/ √ a + 1/ √ c a, b, c are in AP √ a, √ b, √ c are in AP plz explain too

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Aditya Gupta · Answer

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 :))

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Three circles of radii a, b, c (a＜b＜c) touch each other externally. If they have x axis as their common yangent, then: 1/ √ a = 1/ √ b + 1/ √ c 1/ √ b = 1/ √ a + 1/ √ c a, b, c are in AP √ a, √ b, √ c are in AP plz explain too

Three circles of radii a, b, c (a＜b＜c) touch each other externally. If they have x axis as their common yangent, then:

1/√ a = 1/√ b + 1/√ c

1/√ b = 1/√ a + 1/√ c

a, b, c are in AP

√ a, √ b, √ c are in AP
plz explain too

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Three circles of radii a, b, c (a＜b＜c) touch each other externally. If they have x axis as their common yangent, then: 1/ √ a = 1/ √ b + 1/ √ c 1/ √ b = 1/ √ a + 1/ √ c a, b, c are in AP √ a, √ b, √ c are in AP plz explain too

Three circles of radii a, b, c (a＜b＜c) touch each other externally. If they have x axis as their common yangent, then: 1/√ a = 1/√ b + 1/√ c 1/√ b = 1/√ a + 1/√ c a, b, c are in AP √ a, √ b, √ c are in APplz explain too

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Three circles of radii a, b, c (a＜b＜c) touch each other externally. If they have x axis as their common yangent, then:

1/√ a = 1/√ b + 1/√ c

1/√ b = 1/√ a + 1/√ c

a, b, c are in AP

√ a, √ b, √ c are in AP
plz explain too