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 _____________.

The correct answer is 0.84.

Step-by-step Explanation:

1. Understand the Setup and Geometry of Triangle ABC:

We are given a right-angled triangle ABC with:

- Side AB = 1

- Side AC = 3

- $\angle BAC=\frac{\pi }{2}=90°$

Let us set up a Cartesian coordinate system with vertex A at the origin (0, 0):

- Point A = (0, 0)

- Point B = (1, 0) along the positive x-axis

- Point C = (0, 3) along the positive y-axis

2. Circumcircle of Triangle ABC:

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

The midpoint of BC is the center of the circumcircle, O:

$O=\left(\frac{1+0}{2},\frac{0+3}{2}\right)=\left(\frac{1}{2},\frac{3}{2}\right)$

The radius R of the circumcircle is half the length of the hypotenuse BC:

$BC=\sqrt{1^2+3^2}=\sqrt{10}$

$R=\frac{\sqrt{10}}{2}$

3. Equation/Position of the Inner Circle:

A circle of radius r touches the sides AB (x-axis) and AC (y-axis).

Since it lies in the first quadrant touching both axes, its center P must have coordinates (r, r).

4. Condition for Internal Tangency to the Circumcircle:

The inner circle (center P, radius r) touches the circumcircle (center O, radius R) internally.

The condition for two circles to touch internally is that the distance between their centers equals the difference of their radii:

$OP=R-r$

Let us compute the distance OP squared:

$O{P}^2={\left(r-\frac{1}{2}\right)}^2+{\left(r-\frac{3}{2}\right)}^2$

Setting OP equal to R - r and squaring both sides:

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

5. Solving for r:

Expanding the left-hand side:

$\left(r^2-r+\frac{1}{4}\right)+\left(r^2-3r+\frac{9}{4}\right)=2r^2-4r+\frac{10}{4}=2r^2-4r+\frac{5}{2}$

Expanding the right-hand side:

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

Equating both sides:

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

Subtracting $\frac{5}{2}$ and $r^2$ from both sides:

$r^2-4r+\sqrt{10}r=0$

$r\left(r-\left(4-\sqrt{10}\right)\right)=0$

Since r > 0, we have:

$r=4-\sqrt{10}$

6. Calculating the Numerical Value:

Approximating $\sqrt{10}\approx 3.162277$:

$r=4-3.162277=0.837723\approx 0.84$

Thus, the value of r rounded to two decimal places is 0.84.