Step-by-step Derivation:

1. Set up a Coordinate System

   Since the triangle ABC has a right angle at A, it is convenient to place vertex A at the origin (0, 0) of a Cartesian coordinate system. Let the side AB lie along the positive x-axis and the side AC lie along the positive y-axis.

   The coordinates of the vertices are:

   • A = (0, 0)
   • B = (1, 0), since AB = 1.
   • C = (0, 3), since AC = 3.

2. Find the Properties of the Circumcircle of Triangle ABC

   For a right-angled triangle, the circumcenter is the midpoint of the hypotenuse, and the circumradius is half the length of the hypotenuse.

   • The hypotenuse is the side BC. Its length can be found using the distance formula: $BC = \sqrt{(1-0)^2 + (0-3)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1+9} = \sqrt{10}$
   • The radius of the circumcircle, let's call it R, is: $R = \frac{BC}{2} = \frac{\sqrt{10}}{2}$
   • The center of the circumcircle, let's call it $O_c$, is the midpoint of BC: $O_c = \left( \frac{1+0}{2}, \frac{0+3}{2} \right) = \left( \frac{1}{2}, \frac{3}{2} \right)$

3. Find the Properties of the Inner Circle

   Let the inner circle have radius $r$ and its center be $O_s = (h, k)$.

   • The circle touches the side AB, which is the line $y=0$. The distance from the center $(h, k)$ to this line is $|k|$. Since the circle is in the first quadrant (bounded by the triangle sides), $k>0$. This distance must be equal to the radius $r$. So, $k=r$.
   • The circle touches the side AC, which is the line $x=0$. The distance from the center $(h, k)$ to this line is $|h|$. Similarly, $h>0$, so this distance is $h$. This must also be equal to the radius $r$. So, $h=r$.
   • Therefore, the center of the inner circle is $O_s = (r, r)$.

4. Apply the Condition of Internal Tangency

   The problem states that the inner circle touches the circumcircle internally. When two circles touch internally, the distance between their centers is equal to the difference of their radii.

   • Distance between centers, $d(O_c, O_s) = R - r$.
   • We can also express the square of this distance using the distance formula for $O_c = (1/2, 3/2)$ and $O_s = (r, r)$: $d(O_c, O_s)^2 = \left(r - \frac{1}{2}\right)^2 + \left(r - \frac{3}{2}\right)^2$

5. Formulate and Solve the Equation for r

   We set up the equation using the tangency condition: $d(O_c, O_s)^2 = (R-r)^2$ $\left(r - \frac{1}{2}\right)^2 + \left(r - \frac{3}{2}\right)^2 = \left(\frac{\sqrt{10}}{2} - r\right)^2$

   Now, we expand both sides of the equation:

   • Left Hand Side (LHS): $LHS = \left(r^2 - r + \frac{1}{4}\right) + \left(r^2 - 3r + \frac{9}{4}\right) = 2r^2 - 4r + \frac{10}{4} = 2r^2 - 4r + \frac{5}{2}$
   • Right Hand Side (RHS): $RHS = \left(\frac{\sqrt{10}}{2}\right)^2 - 2\left(\frac{\sqrt{10}}{2}\right)r + r^2 = \frac{10}{4} - \sqrt{10}r + r^2 = \frac{5}{2} - \sqrt{10}r + r^2$

   Equating the LHS and RHS: $2r^2 - 4r + \frac{5}{2} = r^2 - \sqrt{10}r + \frac{5}{2}$

   Subtract $\frac{5}{2}$ from both sides and rearrange the terms: $2r^2 - r^2 - 4r + \sqrt{10}r = 0$ $r^2 - (4 - \sqrt{10})r = 0$ $r(r - (4 - \sqrt{10})) = 0$

   This equation gives two possible solutions: $r=0$ or $r = 4 - \sqrt{10}$. The problem states that $r>0$, so we must choose the non-zero solution. $r = 4 - \sqrt{10}$

6. Calculate the Numerical Value

   To find the numerical value of $r$, we use an approximation for $\sqrt{10}$. We know that $3.1^2 = 9.61$ and $3.2^2 = 10.24$. A more accurate value is $\sqrt{10} \approx 3.16227...$ $r = 4 - \sqrt{10} \approx 4 - 3.16227 = 0.83773$ Rounding to three decimal places, we get $r \approx 0.838$. This value lies within the given range of 0.82 to 0.86.