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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
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B
1b=1a=1c
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C
a, b, c are in A.P.
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D
a,b,c are in A.P.
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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 $$\sqrt{abc}$$

$$\Rightarrow\,\dfrac{\sqrt{bc}}{\sqrt{abc}}=\dfrac{\sqrt{ba}}{\sqrt{abc}}+\dfrac{\sqrt{ac}}{\sqrt{abc}}$$

$$\Rightarrow\,\dfrac{1}{\sqrt{a}}=\dfrac{1}{\sqrt{c}}+\dfrac{1}{\sqrt{b}}$$


1490432_1567301_ans_71e1c29b7a104d6caa0a13d5e8e4e16d.PNG

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