Animated Solution for Mathematics - Circles: If a circle C passing through the point (4,0) touches the circle $x^2+y^2+4x-6y=12$ externally at the point (1, -1), then the radius of C is :

A Journey into Circles

Welcome, future engineer! Today, we are going to peel back the layers of a beautiful geometry problem. In the world of JEE Advanced, geometry isn't just about drawing shapes; it is about understanding the hidden relationships between them.

We are given a circle, $C_1$, and we need to find another circle, $C$, that touches it externally at a specific point. This is a classic problem that tests your ability to use the 'Family of Circles'—a concept that is as powerful as it is elegant.

Analyzing the Known World

First, let us look at our given circle, $C_1$. Its equation is $x^2+y^2+4x-6y-12=0$.

Before we do anything else, we must understand its anatomy. By comparing this to the general form $x^2+y^2+2gx+2fy+c=0$, we identify the center as $(-g,-f)=(-2,3)$.

The radius, calculated via $r_1=\sqrt{g^2+f^2-c}$, gives us:

So, we have a circle centered at $(-2,3)$ with a radius of $5$. This is our anchor point.

The Kissing Point

We are told that our unknown circle $C$ touches $C_1$ externally at the point $P(1,-1)$. Imagine these two circles 'kissing' at this exact coordinate.

This point $P$ is the bridge between the known and the unknown. Because they touch at $P$, they share a common tangent line, $L$. If we can find the equation of this line $L$, we hold the key to the entire problem.

The Magic of $T=0$

How do we find the tangent line $L$ without getting lost in slopes and derivatives? We use the $T=0$ formula. This is a beautiful piece of algebraic machinery.

For any circle $S=0$, the tangent at $(x_1,y_1)$ is found by replacing $x^2$ with $x{x}_1$, $y^2$ with $y{y}_1$, $2x$ with $(x+x_1)$, and $2y$ with $(y+y_1)$. Applying this to our circle $C_1$ at point $P(1,-1)$, we get:

Expanding this, we get $x-y+2x+2-3y+3-12=0$. Simplifying, we arrive at the equation of our common tangent line $L$:

This line is the boundary that both circles respect.

The Family of Circles

Now, we invoke the 'Family of Circles' theorem. Any circle that touches $C_1$ at point $P$ can be expressed as $S_1+\lambda L=0$.

This is the most efficient way to handle this problem. We don't need to find the center of the new circle yet; we just need to find the parameter $\lambda$ that defines our specific circle $C$. Our equation becomes:

Solving for the Unknown

We are given one final clue: the circle $C$ passes through the point $A(4,0)$. This point must satisfy our family equation. Let us substitute $x=4$ and $y=0$ into the equation:

Solving for $\lambda$, we find $\lambda =-4$. The mystery is unraveling!

The Final Revelation

Substituting $\lambda =-4$ back into our family equation, we get:

This is the equation of our circle $C$. To find the radius, we compare it to the general form. Here, $g=-4$, $f=5$, and $c=16$. The radius $R$ is:

And there we have it! The radius is 5. Notice how the math guided us perfectly to the answer. The beauty of this method is that it bypasses the need for complex geometric construction, relying instead on the elegant symmetry of the family of circles.