Animated Solution for Mathematics - Circles: Let $r_1$ and $r_2$ be the radii of the largest and smallest circles, respectively, which pass through the point $(-4,1)$ and having their centres on the circumference of the circle $x^2+y^2+2x+4y-4=0$. If $\frac{r_1}{r_2}=a+b\sqrt{2}$, then $a+b$ is equal to :

A Geometric Journey

Welcome, fellow traveler of the coordinate plane. Today, we are not just solving a problem; we are choreographing a dance between a fixed point and a circle.

Imagine you are standing on a vast, infinite grid. You have a fixed circle, a blue ring of possibilities, and a single, lonely point $P$ at $(-4,1)$.

We are tasked with drawing new circles that pass through $P$, but with a catch: their centers must be anchored to the circumference of our blue circle. This is a problem of optimization, of finding the extremes in a sea of possibilities.

Before we can dance, we must know the stage. We are given the equation:

To understand this circle, we must peel back its layers. We compare this to the general form $x^2+y^2+2gx+2fy+c=0$.

By identifying the coefficients, we find the center $C$ at $(-g,-f)$, which gives us $C(-1,-2)$.

Now, what about its size? The radius $R$ is calculated via the formula $R=\sqrt{g^2+f^2-c}$.

Plugging in our values, we get:

We now have a circle centered at $(-1,-2)$ with a radius of $3$. This is our anchor.

Now, let us connect our point $P(-4,1)$ to the center $C(-1,-2)$. This distance, $CP$, is the bridge that will allow us to reach the extreme radii.

Using the distance formula $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$, we calculate the gap between these two points.

The difference in $x$ is $-4-(-1)=-3$, and the difference in $y$ is $1-(-2)=3$. Squaring these gives us $9$ and $9$, respectively.

Thus, the distance is:

This value, $3\sqrt{2}$, is the heartbeat of our problem.

Here is where the magic happens. The radius of any new circle we draw is simply the distance from its center (which lies on our blue circle) to the point $P$.

To make this radius as large as possible, we need to push the center as far away from $P$ as possible. To make it as small as possible, we pull the center as close to $P$ as possible.

Geometry tells us that the extreme distances from a point to a circle always lie along the line connecting the point to the center of the circle. Imagine a line passing through $P$ and $C$. It intersects the circle at two points: one is the 'farthest' point, and the other is the 'closest' point.

Therefore, the maximum radius $r_1$ is $CP+R$, and the minimum radius $r_2$ is $CP-R$. Substituting our values:

The problem asks for the ratio $\frac{r_1}{r_2}$. Let us set this up:

We must rationalize the denominator by multiplying the numerator and the denominator by the conjugate, $(\sqrt{2}+1)$.

The denominator becomes $(\sqrt{2}{)}^2-1^2=2-1=1$. The numerator becomes $(\sqrt{2}+1{)}^2$.

Expanding this using the identity $(a+b{)}^2=a^2+b^2+2ab$, we get:

The Victory

We have arrived at the form $a+b\sqrt{2}=3+2\sqrt{2}$. By simple comparison, we see that $a=3$ and $b=2$.

The final step is to find $a+b$, which is:

Look at what you have achieved! You navigated the coordinate plane, utilized the properties of circles, mastered the distance formula, and performed algebraic rationalization with precision. This is the essence of JEE Advanced mathematics—not just memorizing formulas, but understanding the geometric soul of the problem.