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

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$a, b, c are in A.P.$

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

$\sqrt{a}, \sqrt{b}, \sqrt{c}$ are in A.P.

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

answer is A.

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Detailed Solution

Explanation:

We have three circles with radii a, b, and c (where a < b < c), and they touch each other externally. The x-axis is a common tangent for all three circles. We need to find the relationship between the radii a, b, and c.

Step-by-Step Analysis

For three circles touching externally, the distance between the centers of two circles is equal to the sum of their radii.

Let’s denote the centers of the three circles as O₁, O₂, and O₃ (for circles with radii a, b, and c respectively). The distances between the centers will be:

Distance between O₁ and O₂: a + b
Distance between O₂ and O₃: b + c
Distance between O₁ and O₃: a + c

Key Observation

Since the circles touch each other externally and share the x-axis as a common tangent, the sum of the reciprocals of the square roots of the radii follows a specific relationship.

Using geometry and the tangency condition between the circles and the x-axis, it can be shown that:

1/√a = 1/√b + 1/√c

Final Answer

Thus, Option A is the correct answer: 1/√a = 1/√b + 1/√c

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