Animated Solution for Mathematics - Circles: The two circles $x^2+y^2=ax$ and $x^2+y^2=c^2$ ($c>0$) touch each other if

Imagine you are standing on a coordinate plane, looking at two circles. One is anchored firmly at the origin, a perfect, symmetric loop. The other is a bit more dynamic, shifting its position based on a parameter $a$.

The question is simple yet profound: under what condition do these two shapes kiss, just touching at a single point? This is not just an algebra problem; it is a dance of geometry.

First, let us look at our players. The second circle, $S_2$, is defined by $x^2+y^2=c^2$. This is the classic, standard circle, centered at the origin $(0,0)$ with a radius $r_2=c$.

Now, consider the first circle, $S_1$, given by $x^2+y^2-ax=0$. To understand its soul, we must transform it. By grouping the $x$ terms and completing the square, we rewrite it as $(x^2-ax)+y^2=0$.

Adding $(\frac{a}{2}{)}^2$ to both sides, we get:

This reveals that $S_1$ is centered at $C_1=(\frac{a}{2},0)$ with a radius $r_1=|\frac{a}{2}|$. Remember, the absolute value is vital here because a radius is a physical distance, and distance cannot be negative.

Distance Between Centers

Now, we need to bridge the gap between these two circles. The distance $d$ between their centers $C_1(\frac{a}{2},0)$ and $C_2(0,0)$ is simply the difference in their $x$-coordinates:

This distance is the key to everything. When two circles touch, they do so in one of two ways: externally or internally.

For external touching, the distance between centers must be the sum of the radii: $d=r_1+r_2$. For internal touching, the distance must be the absolute difference of the radii: $d=|r_1-r_2|$.

Let us test the external case first. If $d=r_1+r_2$, then:

Subtracting $|\frac{a}{2}|$ from both sides leaves us with $c=0$. But the problem states $c>0$. This contradiction tells us that external touching is impossible.

So, we turn to the internal case: $d=|r_1-r_2|$. Substituting our values, we get:

Since the first circle is inside the second, $c$ must be larger than $|\frac{a}{2}|$. Thus, the equation becomes:

Adding $|\frac{a}{2}|$ to both sides, we get $2|\frac{a}{2}|=c$, which simplifies beautifully to $|a|=c$.

There it is. The condition for these two circles to touch is simply $|a|=c$. It is a result of striking simplicity, emerging from the interplay of geometry and algebra.

Whether $a$ is positive or negative, the magnitude of $a$ must match the radius of the larger circle. This is the beauty of JEE mathematics—taking a complex-looking problem and finding the elegant, underlying truth.