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 :

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

When three circles of radii a, b, c touch each other externally, such that all X-axis is their common tangent, then $\frac{1}{\sqrt{a} } = \frac{1}{\sqrt{b} } + \frac{1}{\sqrt{c} }$.

Step-by-step explanation:

The possible figure for given conditions, is as follows.

• Radii $a$ less than $b$ and $b$ is less than $c$.
• Common tangent is X- axis.

Thus, from the figure below:

$CB = CA + AB$

where,

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

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

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

Thus on substituting,

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

On expansion,

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

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

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

For easy simplification, the whole equation is divided by $\sqrt{abc}$

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

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

on re-arranging,

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

Thus, for the given conditions, the relation between three circles radii is given as, $\frac{1}{\sqrt{a} } = \frac{1}{\sqrt{b} } + \frac{1}{\sqrt{c} }$.