Animated Solution for Mathematics - Circles: The equation of the line passing through the points of intersection of the circles $3x^2+3y^2-2x+12y-9=0$ and $x^2+y^2+6x+2y-15=0$ is .........

Finding the Common Chord

Welcome, fellow explorers of the Cartesian plane! Today, we are going to unravel a classic problem in coordinate geometry. We have two circles, $S_1$ and $S_2$, and we want to find the equation of the line that passes through their points of intersection.

This line is known as the common chord. Imagine these two circles sitting on your graph paper; the common chord is the bridge connecting their intersection points. It is a simple, elegant line, but finding it requires a precise touch.

A Powerful Tool

The secret to solving this problem lies in the concept of the radical axis. The radical axis of two circles is the locus of points from which the tangents drawn to both circles have equal lengths.

Mathematically, the equation of this line is given by $S_1-S_2=0$. It is a beautiful piece of algebra where the quadratic terms—the $x^2$ and $y^2$—vanish, leaving us with a linear equation.

But, as with all powerful tools, there is a catch. You cannot simply subtract the equations as they are given. You must ensure that the coefficients of $x^2$ and $y^2$ in both equations are exactly $1$.

This is the normalization step, and it is where most students stumble. If you skip this, you are not finding a line; you are finding a curve, and your answer will be incorrect.

Normalizing the Circles

Let us look at our given circles:

Notice that $S_1$ has coefficients of $3$ for $x^2$ and $y^2$. We must normalize this. Dividing the entire equation by $3$, we get:

Now, look at $S_2$. The coefficients are already $1$. Perfect! We can keep it as it is:

With both equations normalized, we are ready to perform the subtraction. This is the moment where the quadratic terms will cancel out, leaving us with the linear equation of the common chord.

The Algebraic Dance

Now, we subtract equation $(1)$ from equation $(2)$:

Watch closely as the $x^2$ and $y^2$ terms vanish into thin air. Now, let us group the remaining terms.

For the $x$ terms, we have $6x-(-\frac{2}{3}x)$, which simplifies to $6x+\frac{2}{3}x=\frac{18+2}{3}x=\frac{20}{3}x$. For the $y$ terms, we have $2y-4y=-2y$. Finally, for the constants, we have $-15-(-3)=-15+3=-12$.

Putting it all together, we get:

To make this equation look cleaner, let us multiply the entire equation by $3$ to clear the fraction:

We are almost at the finish line! Notice that all coefficients are even. We can simplify this by dividing by $2$:

And there it is! This is the equation of our common chord. It is a simple, linear equation that perfectly describes the line passing through the intersection of our two circles.

Remember, the key was the normalization. Without that first step, the quadratic terms would have remained, and we would have been lost. Keep this in mind, and you will never be trapped by this problem again. Happy calculating!