Animated Solution for Mathematics - Circles: Let $ABC$ be the triangle with $AB=1,AC=3$ and $\angle BAC=\frac{\pi }{2}$. If a circle of radius $r>0$ touches the sides $AB,AC$ and also touches internally the circumcircle of the triangle $ABC$, then the value of $r$ is _____________.

Setting the Stage

Geometry is often a game of perspective. When you look at a problem like this, the first instinct might be to reach for complex trigonometric identities or circle theorems. But wait—let's pause.

The most powerful tool in our arsenal is often the simplest: the Cartesian coordinate system. By anchoring our triangle $ABC$ at the origin, we turn a daunting geometric puzzle into a structured algebraic journey.

Imagine placing the right angle at $A(0,0)$. Since $AB=1$ and $AC=3$, we immediately define our vertices: $A(0,0)$, $B(1,0)$, and $C(0,3)$.

This isn't just a coordinate setup; it's a strategic choice that simplifies every subsequent calculation. We have transformed the abstract triangle into a concrete set of points on a grid.

A Geometric Gift

Now, consider the circumcircle of this right-angled triangle. There is a beautiful, fundamental property here: the circumcenter of any right-angled triangle lies exactly at the midpoint of its hypotenuse.

Our hypotenuse is the segment $BC$. Using the midpoint formula, the center $C_1$ is simply:

What about the circumradius $R$? It is the distance from the center to any vertex, which is half the length of the hypotenuse $BC$.

The length of $BC$ is $\sqrt{1^2+3^2}=\sqrt{10}$. Thus, $R=\frac{\sqrt{10}}{2}$. We have now unlocked the secrets of the larger circle.

The Internal Constraint

Next, we introduce the smaller circle of radius $r$. It touches $AB$ (the $x$-axis) and $AC$ (the $y$-axis).

Because it is nestled in the corner at the origin, its center $C_2$ must be at $(r,r)$. This is the key to linking the two circles.

The problem states that this smaller circle touches the circumcircle internally. Geometrically, this means the distance between the two centers, $C_1$ and $C_2$, must be exactly the difference of their radii: $d=R-r$.

Now, we apply the distance formula between $C_1\left(\frac{1}{2},\frac{3}{2}\right)$ and $C_2(r,r)$:

I know, this equation looks intimidating with the square root and the fractions. But take a breath. We are just one squaring operation away from clarity.

Squaring both sides gives us:

Expanding the left side, we get $(r^2-r+\frac{1}{4})+(r^2-3r+\frac{9}{4})$, which simplifies to $2r^2-4r+\frac{10}{4}$. On the right side, we expand to get $\frac{10}{4}-\sqrt{10}r+r^2$.

Look closely at the equation now: $2r^2-4r+\frac{10}{4}=\frac{10}{4}-\sqrt{10}r+r^2$. Do you see it? The $\frac{10}{4}$ terms on both sides cancel out completely!

This is the moment where the complexity collapses into elegance. We are left with:

Rearranging this, we get $r^2-(4-\sqrt{10})r=0$. Since $r>0$, we can safely divide by $r$ to find $r=4-\sqrt{10}$.

With $\sqrt{10}\approx 3.162$, our radius $r$ is approximately $4-3.162=0.838$. Rounding to two decimal places, we arrive at $0.84$.

You have successfully navigated the coordinate plane, applied the internal tangency condition, and conquered the algebra. This is the essence of JEE Advanced physics and math: turning a complex visual problem into a series of logical, manageable steps.