Animated Solution for Mathematics - Circles: Let the latus rectum of the parabola $y^2=4x$ be the common chord to the circles $C_1$ and $C_2$ each of them having radius $2\sqrt{5}$. Then, the distance between the centres of the circles $C_1$ and $C_2$ is:

Welcome, students. Today, we are not just solving a problem; we are peeling back the layers of a beautiful geometric puzzle. In the world of JEE Advanced, we often encounter problems that look like algebraic nightmares, but if you pause and visualize the geometry, they transform into elegant, simple truths.

Let us embark on this journey with the parabola $y^2=4x$.

First, let us look at our foundation. The equation $y^2=4x$ is the classic, standard form of a parabola, $y^2=4ax$. By comparing the coefficients, we immediately identify that $4a=4$, which gives us $a=1$.

This value, $a=1$, is the heartbeat of our problem. It tells us exactly where the focus lies and, more importantly, defines the latus rectum.

The latus rectum is the chord passing through the focus, perpendicular to the axis of the parabola. Since $a=1$, the equation of this line is simply $x=1$.

If we look at the endpoints, they are at $(a,2a)$ and $(a,-2a)$, which translates to $(1,2)$ and $(1,-2)$. The total length of this segment is $4a=4$.

This vertical line segment is the common chord for our two circles, $C_1$ and $C_2$. Imagine this line segment as a bridge connecting two worlds—the two circles are anchored to this bridge.

Here is where the magic happens. We are told that $C_1$ and $C_2$ share this chord. A fundamental property of geometry states that the centers of circles sharing a common chord must lie on the perpendicular bisector of that chord.

Since our chord is the vertical line $x=1$, its perpendicular bisector is a horizontal line passing through its midpoint. The midpoint of the segment from $(1,2)$ to $(1,-2)$ is $(1,0)$.

Thus, the perpendicular bisector is the x-axis ($y=0$). This means both centers, $C_1$ and $C_2$, must lie on the x-axis. Let us denote their centers as $(h,0)$.

Now, let us focus on one circle. We know its radius $r=2\sqrt{5}$. We have a chord of length $L=4$.

If we draw a perpendicular from the center $(h,0)$ to the chord $x=1$, the distance $d$ is simply $|h-1|$. This creates a right-angled triangle where the hypotenuse is the radius $r$, one leg is the distance $d$, and the other leg is half the length of the chord, which is $\frac{L}{2}=2$.

By the Pythagorean theorem, we have the beautiful relationship:

Substituting our known values:

We have found that the distance $d$ from the center to the chord is $4$. Since $d=|h-1|$, we have two possibilities:

1. $h-1=4\Rightarrow h=5$ 2. $h-1=-4\Rightarrow h=-3$

So, the centers are located at $(5,0)$ and $(-3,0)$. The distance between these two centers is simply the difference in their x-coordinates:

And there we have it. The distance between the centers is 8. Notice how we didn't need complex coordinate geometry equations for the circles themselves? We used the symmetry of the parabola and the elegance of the Pythagorean theorem. Keep this mindset—always look for the geometric shortcut before diving into the algebra. You have mastered this problem.