Key Concepts and Formulas

• Circles touching externally: If two circles with radii $r_1$ and $r_2$ touch each other externally, the distance between their centers is $r_1 + r_2$.
• Circle tangent to the x-axis: If a circle of radius $r$ is tangent to the x-axis, its center will have coordinates $(x, r)$ for some $x$.
• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Step-by-Step Solution

Step 1: Define the Centers

Let the centers of the circles with radii $a$, $b$, and $c$ be $A$, $B$, and $C$, respectively. Since the circles are tangent to the x-axis, their centers will have coordinates $A(x_1, a)$, $B(x_2, b)$, and $C(x_3, c)$.

Step 2: Apply the Distance Formula and External Tangency

Since the circles touch each other externally, we can apply the distance formula between the centers.

• Distance between A and B: $AB = a + b$ $\sqrt{(x_2 - x_1)^2 + (b - a)^2} = a + b$ Squaring both sides: $(x_2 - x_1)^2 + (b - a)^2 = (a + b)^2$ $(x_2 - x_1)^2 = (a + b)^2 - (b - a)^2 = a^2 + 2ab + b^2 - (b^2 - 2ab + a^2) = 4ab$ $x_2 - x_1 = \pm 2\sqrt{ab}$

• Distance between B and C: $BC = b + c$ $\sqrt{(x_3 - x_2)^2 + (c - b)^2} = b + c$ Squaring both sides: $(x_3 - x_2)^2 + (c - b)^2 = (b + c)^2$ $(x_3 - x_2)^2 = (b + c)^2 - (c - b)^2 = b^2 + 2bc + c^2 - (c^2 - 2bc + b^2) = 4bc$ $x_3 - x_2 = \pm 2\sqrt{bc}$

• Distance between A and C: $AC = a + c$ $\sqrt{(x_3 - x_1)^2 + (c - a)^2} = a + c$ Squaring both sides: $(x_3 - x_1)^2 + (c - a)^2 = (a + c)^2$ $(x_3 - x_1)^2 = (a + c)^2 - (c - a)^2 = a^2 + 2ac + c^2 - (c^2 - 2ac + a^2) = 4ac$ $x_3 - x_1 = \pm 2\sqrt{ac}$

Step 3: Establish a Relationship

Since the circles are tangent to the x-axis in the same order (a < b < c), we can assume without loss of generality that $x_1 < x_2 < x_3$. Therefore, we take the positive roots in the previous step. Then:

• $x_2 - x_1 = 2\sqrt{ab}$
• $x_3 - x_2 = 2\sqrt{bc}$
• $x_3 - x_1 = 2\sqrt{ac}$

Also, $(x_3 - x_1) = (x_3 - x_2) + (x_2 - x_1)$. Substituting the values, we get:

$2\sqrt{ac} = 2\sqrt{bc} + 2\sqrt{ab}$

Dividing by 2, we get:

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

Dividing by $\sqrt{abc}$, we get:

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

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

Step 4: Check the Options

Comparing with the given options, we see that option (D) matches our result.

Common Mistakes & Tips

• Always remember that the distance between the centers of two circles touching externally is the sum of their radii.
• When taking square roots, consider both positive and negative values. However, in this case, the order of tangency on the x-axis restricts the sign.
• Simplifying the equation by dividing by a common factor (in this case, $\sqrt{abc}$) is a useful technique.

Summary

By setting up the coordinates of the circle centers and using the distance formula, along with the condition that the circles touch externally and the x-axis, we derived the relationship $\frac{1}{\sqrt{b}} = \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{c}}$. This corresponds to option (D).

Final Answer The final answer is \boxed{D}, which corresponds to option (D).