Animated Solution for Mathematics - Circles: Sketch the solution set of the following system of inequalities: $x^2+y^2-2x\le 0;3x-y-12\le 0;y-x\le 0;y\ge 0$.

Welcome, future engineers! Today, we are not just solving a system of inequalities; we are embarking on a cartographic expedition. Imagine you are standing in a coordinate plane, and you have been given a set of four rules—four 'clues'—that define a specific, hidden region.

Our goal is to sketch this region and find the exact territory where all these conditions coexist in harmony. This is the essence of linear and non-linear programming, a cornerstone of optimization theory that you will use throughout your engineering careers.

Our first clue is the most intimidating: $x^2+y^2-2x\le 0$. When you see $x^2$ and $y^2$ with equal coefficients, your intuition should immediately scream, "Circle!"

To make sense of it, we need to bring it into the standard form of a circle: $(x-h{)}^2+(y-k{)}^2=r^2$. We use the technique of 'completing the square' by grouping the $x$ terms: $(x^2-2x)+y^2\le 0$.

To make $(x^2-2x)$ a perfect square, we add $1$ to both sides of the inequality to maintain balance. This transforms our equation into:

Suddenly, the fog clears! We have a circle centered at $(1,0)$ with a radius of $1$. The inequality $\le 1$ tells us that our treasure is not just on the boundary, but anywhere inside this circular disk.

Now, let's look at the linear constraints. The second clue is $y-x\le 0$, which simplifies to $y\le x$. This is a line passing through the origin at a $45^{\circ }$ angle.

To decide which side to shade, we use the test point method. If we pick a point like $(1,0)$, we see that $0\le 1$ is true. Therefore, we shade the region below the line $y=x$.

Our third clue is $y\ge 0$. This is the x-axis itself, acting as a floor that prevents our solution from dipping into the negative y-territory. We are now restricted to the upper half-plane.

Finally, we encounter the fourth clue: $3x-y-12\le 0$. This can be rewritten as $y\ge 3x-12$.

Before you start drawing lines, think about the geometry. Our circle is trapped between $x=0$ and $x=2$. The line $y=3x-12$ crosses the x-axis at $x=4$, meaning it lies entirely to the right of our circle.

Since the entire circle satisfies $y\ge 3x-12$, this is a redundant constraint. It is a red herring designed to test your ability to analyze the system before blindly calculating. We can safely ignore it.

We are left with the intersection of the circle $(x-1{)}^2+y^2\le 1$, the line $y\le x$, and the condition $y\ge 0$. To find the exact vertices of our region, we solve for the intersection of the circle and the line $y=x$.

Substituting $y=x$ into the circle equation gives:

Expanding this, we get $x^2-2x+1+x^2=1$, which simplifies to $2x^2-2x=0$. Factoring this, we find $2x(x-1)=0$, giving us $x=0$ and $x=1$.

This tells us that the line $y=x$ intersects the circle at the origin $(0,0)$ and the point $(1,1)$.

Putting it all together, our solution set is the circular segment bounded by the origin $(0,0)$, the point $(1,1)$, and the rightmost edge of the circle at $(2,0)$. It is a beautiful, curved slice of the plane.

By breaking down the complex system into individual, manageable geometric shapes, we have successfully mapped the solution. Remember, in JEE Advanced, the math is just the language; the geometry is the story. Keep visualizing, keep testing, and keep exploring!