Radius of Circle Touching Triangle Sides and Circumcircle Internally

To find the radius $r > 0$ of the circle, we set up a Cartesian coordinate system with vertex $A$ as the origin.

Step 1: Coordinate Setup

• Place vertex $A$ at the origin $(0, 0)$.
• Since $\angle BAC = \frac{\pi}{2}$, let side $AB$ lie along the positive $x$-axis and side $AC$ lie along the positive $y$-axis.
• Given $AB = 1$ and $AC = 3$, the coordinates of the vertices are: $A = (0, 0), \quad B = (1, 0), \quad C = (0, 3)$

Step 2: Circumcircle of $\triangle ABC$

Since $\triangle ABC$ is a right-angled triangle at $A$, the hypotenuse $BC$ is the diameter of its circumcircle.

• Circumcenter ($O$): The midpoint of $BC$: $O = \left(\frac{1 + 0}{2}, \frac{0 + 3}{2}\right) = \left(\frac{1}{2}, \frac{3}{2}\right)$
• Circumradius ($R$): Half the length of the hypotenuse $BC$: $R = \frac{1}{2}\sqrt{1^2 + 3^2} = \frac{\sqrt{10}}{2}$

Step 3: Equation for the Required Circle

Let $O'$ be the center of the circle of radius $r$.

• Since this circle touches the line $AB$ ($y = 0$) and the line $AC$ ($x = 0$) in the first quadrant, its center is: $O' = (r, r)$

Step 4: Using the Tangency Condition

The circle touches the circumcircle of $\triangle ABC$ internally. Therefore, the distance between their centers $O$ and $O'$ is equal to the difference of their radii: $OO' = R - r$

Using the distance formula: $\sqrt{\left(r - \frac{1}{2}\right)^2 + \left(r - \frac{3}{2}\right)^2} = \frac{\sqrt{10}}{2} - r$

Squaring both sides: $\left(r - \frac{1}{2}\right)^2 + \left(r - \frac{3}{2}\right)^2 = \left(\frac{\sqrt{10}}{2} - r\right)^2$

Expanding each side: $\left(r^2 - r + \frac{1}{4}\right) + \left(r^2 - 3r + \frac{9}{4}\right) = \frac{10}{4} - \sqrt{10}r + r^2$

$2r^2 - 4r + \frac{5}{2} = r^2 - \sqrt{10}r + \frac{5}{2}$

Subtracting $r^2 + \frac{5}{2}$ from both sides gives: $r^2 - 4r + \sqrt{10}r = 0$ $r\left(r - 4 + \sqrt{10}\right) = 0$

Since $r > 0$, we get: $r = 4 - \sqrt{10}$

Verification of Validity

• Since $\sqrt{10} \approx 3.162$, $r = 4 - 3.162 = 0.838 > 0$.
• The tangency point on side $AB$ is $(r, 0) \approx (0.838, 0)$, which lies on the segment $AB$ as $0 < 0.838 < 1$.
• The tangency point on side $AC$ is $(0, r) \approx (0, 0.838)$, which lies on the segment $AC$ as $0 < 0.838 < 3$.

Thus, the exact value of $r$ is $4 - \sqrt{10}$ (or approximately $0.84$).