Question

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

A
1a=1b+1c
B
1b=1a=1c
C
a, b, c are in A.P.
D
a,b,c are in A.P.

Solution

## The correct option is A $$\dfrac{1}{\sqrt{a}}=\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}$$$$BC=AC+AB$$$$\Rightarrow\,\sqrt{{\left(b+c\right)}^{2}-{\left(b-c\right)}^{2}}=\sqrt{{\left(b+a\right)}^{2}-{\left(b-a\right)}^{2}}+\sqrt{{\left(a+c\right)}^{2}-{\left(a-c\right)}^{2}}$$$$\sqrt{{b}^{2}+{c}^{2}+2bc-{b}^{2}-{c}^{2}+2bc}=\sqrt{{b}^{2}+{a}^{2}+2ba-{b}^{2}-{a}^{2}+2ba}+\sqrt{{a}^{2}+{c}^{2}+2ac-{a}^{2}-{c}^{2}+2ca}$$$$\Rightarrow\,\sqrt{4bc}=\sqrt{4ba}+\sqrt{4ac}$$$$\Rightarrow\,2\sqrt{bc}=2\sqrt{ba}+2\sqrt{ac}$$$$\Rightarrow\,\sqrt{bc}=\sqrt{ba}+\sqrt{ac}$$Divide both sides by $$\sqrt{abc}$$$$\Rightarrow\,\dfrac{\sqrt{bc}}{\sqrt{abc}}=\dfrac{\sqrt{ba}}{\sqrt{abc}}+\dfrac{\sqrt{ac}}{\sqrt{abc}}$$$$\Rightarrow\,\dfrac{1}{\sqrt{a}}=\dfrac{1}{\sqrt{c}}+\dfrac{1}{\sqrt{b}}$$Maths

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