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

Answer 1

The correct option is A $\frac{1}{\sqrt{a}}=\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}$

$BC=AC+AB$

$\Rightarrow \sqrt{{(b+c)}^2-{(b-c)}^2}=\sqrt{{(b+a)}^2-{(b-a)}^2}+\sqrt{{(a+c)}^2-{(a-c)}^2}$

$\sqrt{b^2+c^2+2bc-b^2-c^2+2bc}=\sqrt{b^2+a^2+2ba-b^2-a^2+2ba}+\sqrt{a^2+c^2+2ac-a^2-c^2+2ca}$

$\Rightarrow \sqrt{4bc}=\sqrt{4ba}+\sqrt{4ac}$

$\Rightarrow 2\sqrt{bc}=2\sqrt{ba}+2\sqrt{ac}$

$\Rightarrow \sqrt{bc}=\sqrt{ba}+\sqrt{ac}$

Divide both sides by $\sqrt{abc}$

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

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