Animated Solution for Mathematics - Circles: The points of intersection of the line $4x-3y-10=0$ and the circle $x^2+y^2-2x+4y-20=0$ are ......... and .........

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are witnessing a beautiful dance between geometry and algebra.

We have a circle, defined by the equation $x^2+y^2-2x+4y-20=0$, and a line, defined by $4x-3y-10=0$. Our mission is to find the exact points where this line slices through the circle.

The points where they meet are the only places in the entire plane where the line and the circle share the exact same $(x,y)$ coordinates. This is the core of our problem: finding the common ground.

To find these points, we need a strategy. We cannot simply look at the graph and guess; we need the precision of algebra.

The most robust tool in our arsenal is the method of substitution. We treat the linear equation as a bridge. By isolating one variable, we can express the entire line in terms of the other variable.

Let's take our line $4x-3y-10=0$ and isolate $x$. Moving the terms, we get $4x=3y+10$, which simplifies to:

This is our key. It allows us to translate any point on the line into a coordinate pair that depends only on $y$.

Now, we take this expression for $x$ and substitute it into the circle's equation. We replace every $x$ in $x^2+y^2-2x+4y-20=0$ with $\frac{3y+10}{4}$.

The equation becomes:

It looks intimidating, but it is just a quadratic equation in disguise. By expanding the squared term, we get:

To make our lives easier, we multiply the entire equation by $16$ to clear the denominators. This leaves us with:

Now, we combine like terms. Grouping the $y^2$ terms, we get $25y^2$. The $y$ terms combine to $60y-24y+64y=100y$, and the constants combine to $100-80-320=-300$.

Our equation is now:

Dividing by $25$, we arrive at the elegant:

This is a simple quadratic that factors beautifully into $(y+6)(y-2)=0$. Thus, our intersection points occur at $y=2$ and $y=-6$.

Substituting these back into our bridge equation, $x=\frac{3y+10}{4}$, we find the corresponding $x$ values:

For $y=2$, $x=\frac{3(2)+10}{4}=4$.

For $y=-6$, $x=\frac{3(-6)+10}{4}=-2$.

The intersection points are $(4,2)$ and $(-2,-6)$. Algebra has perfectly predicted the geometry. Keep practicing, and you will see this elegance everywhere.