Animated Solution for Mathematics - Circles: 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 Visual Journey

Imagine standing on a flat, infinite plane—the $x$-axis. You are watching three circles of varying sizes, with radii $a$, $b$, and $c$, rolling toward each other until they kiss perfectly, touching externally.

They are all resting on this same $x$-axis. We are given that $a<b<c$, which tells us that the smallest circle, $a$, is the one caught in the middle, wedged between the two larger giants, $b$ and $c$.

The Tangent Formula

Before we dive into the algebra, we need to equip ourselves with a powerful geometric tool. When two circles of radii $r_1$ and $r_2$ touch externally, we must determine the distance between their contact points on the common tangent.

If you draw the centers of these circles, you form a right-angled triangle. The hypotenuse is the sum of the radii, $(r_1+r_2)$, and one side is the difference, $|r_1-r_2|$.

By the Pythagorean theorem, the distance $L$ between the contact points is:

When you expand this, the $r_1^2$ and $r_2^2$ terms cancel out, leaving you with $\sqrt{4r_1{r}_2}$, which simplifies beautifully to:

This is the key that unlocks the entire problem.

Now, let's look at our configuration. The distance between the contact points of the largest circle $c$ and the middle circle $b$ is $D_{bc}=2\sqrt{bc}$.

Similarly, the distance between the smallest circle $a$ and the largest circle $c$ is $D_{ac}=2\sqrt{ac}$. Finally, the distance between the smallest circle $a$ and the middle circle $b$ is $D_{ab}=2\sqrt{ab}$.

Because the smallest circle $a$ is wedged perfectly between $b$ and $c$, the total distance between the contact points of $b$ and $c$ must be the sum of the distances between the contact points of $a$ and $c$, and $a$ and $b$. Mathematically, we write this as:

This is where the elegance of the solution shines. First, we divide the entire equation by $2$, leaving us with $\sqrt{bc}=\sqrt{ac}+\sqrt{ab}$.

Now, to isolate the variables and find the relationship between the radii, we divide every term by $\sqrt{abc}$:

Look closely at the terms. The common factors cancel out, leaving us with the final, elegant result:

This is the beauty of JEE Advanced mathematics. We started with a complex visual arrangement of circles, applied a fundamental geometric theorem, and through a series of logical steps, arrived at a clean, symmetric algebraic relationship.