Animated Solution for Mathematics - Circles: Let $ABCD$ and $AEFG$ be squares of side 4 and 2 units, respectively. The point $E$ is on the line segment $AB$ and the point $F$ is on the diagonal $AC$. Then the radius $r$ of the circle passing through the point $F$ and touching the line segments $BC$ and $CD$ satisfies:

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are uncovering a hidden structure.

Geometry is the language of the universe, and coordinate geometry is the translator that allows us to speak it fluently. Let us dive into this problem of squares and circles, where we will transform a visual puzzle into a beautiful algebraic dance.

Establishing the Foundation

Imagine you are standing at point $A$, the origin $(0,0)$ of our coordinate system. Before us lies the square $ABCD$, a grand structure with a side length of 4 units.

By placing $A$ at the origin, we immediately unlock the coordinates of the other vertices: $B$ at $(4,0)$, $C$ at $(4,4)$, and $D$ at $(0,4)$. This is our canvas.

Now, consider the smaller square $AEFG$ with a side length of 2 units. Since $E$ lies on $AB$, it sits at $(2,0)$. The point $F$ is the key—it lies on the diagonal $AC$. Since the diagonal of a square starting from the origin follows the line $y=x$, and $F$ is a vertex of the square $AEFG$ with side 2, its coordinates must be $(2,2)$. We have now mapped the entire landscape.

The Tangency Trap

Now, we introduce the circle. It is not just floating; it is constrained. It touches the line segments $BC$ (which is the line $x=4$) and $CD$ (which is the line $y=4$).

This is the crucial moment. If a circle of radius $r$ is trapped in the corner of a square, its center cannot be anywhere. It must be exactly $r$ units away from the boundaries.

If we move $r$ units to the left from $x=4$, we get $x=4-r$. If we move $r$ units down from $y=4$, we get $y=4-r$. Thus, the center of our circle is $O(4-r,4-r)$. This is the heartbeat of our solution.

The Bridge of Distance

We know the circle passes through point $F(2,2)$. In the world of coordinate geometry, the distance from the center of a circle to any point on its circumference is always the radius $r$.

We have our center $(4-r,4-r)$ and our point $F(2,2)$. Let us invoke the distance formula, the bridge that connects these two points:

Substituting our values, we get:

This looks intimidating, but let us breathe. Simplify the terms inside the brackets: $2-4+r=r-2$. The equation becomes:

This is the moment of clarity. We have two identical terms, so we can combine them:

The Algebraic Climax

Now, we expand the perfect square. Remember that $(r-2{)}^2=r^2-4r+4$. Multiplying this by 2, we get:

Distributing the 2, we arrive at:

Finally, we bring everything to one side to set the equation to zero. Subtracting $r^2$ from both sides, we find:

Look at what we have achieved. We started with a geometric description, translated it into coordinates, applied the logic of tangency, and used the distance formula to derive a quadratic equation.

The result, $r^2-8r+8=0$, is not just an answer; it is the mathematical signature of the circle's existence within that square. You have successfully navigated the problem. Keep this mindset—break the complex into the simple, and the solution will always reveal itself.