Find the radius of the inner circle

Let the center of the orange circle be $O$ , let the center of the blue circle be $P$ , let the radius of the blue circle be $x$ , and label the diagram as follows:

By Thales's Theorem, $\angle ACB$ is a right angle, and by the Pythagorean Theorem, $AB = \sqrt{AC^2 + CB^2} = \sqrt{12^2 + 5^2} = 13$ , so $AO = OB = \frac{13}{2}$ , which makes $OP = OI - PI = \frac{13}{2} - x$ .

As a tangent to the blue circle, $\angle PDC$ and $\angle BEC$ are also right angles, so $PDCE$ is a rectangle, and since $PD = PE = x$ , $PDCE$ is a square, so $CD = CE = x$ as well, so $AD = AC - CD = 12 - x$ .

Now $\triangle ACB \sim \triangle ADG \sim \triangle HPG$ by AA similarity, so $\cos \angle OGP = - \cos \angle AGD = -\cos \angle ABC = -\frac{5}{13}$ , and $AG = \frac{AB}{AC} \cdot AD = 13 - \frac{13}{12}x$ and $DG = \frac{BC}{AC} \cdot AD = 5 - \frac{5}{12}x$ , which means $OG = AG - AO = 13 - \frac{13}{12}x - \frac{13}{2} = \frac{13}{2} - \frac{13}{12}x$ and $GP = DP - DG = x - (5 - \frac{5}{12}x) = \frac{17}{12}x - 5$ .

Then by the law of cosines on $\triangle OGP$ , $\cos \angle OGP = \cfrac{OG^2 + GP^2 - OP^2}{2 \cdot OG \cdot GP}$ , or after substituting, $-\cfrac{5}{13} = \cfrac{(\frac{13}{2} - \frac{13}{12}x)^2 + (\frac{17}{12}x - 5)^2 - (\frac{13}{2} - x)^2}{2 \cdot (\frac{13}{2} - \frac{13}{12}x) \cdot (\frac{17}{12}x - 5)}$ , which solves to $x = \boxed{4}$ for $x > 0$ .

Geogebra video

1. Find incentrum $E$ of triangle $ABC$ .
2. Find point $F$ intersection circle and middle perpendicular $BC$ .
3. Find point $G$ intersection circle and line $FE$ .
4. Find point $H$ of intersection of line $GD$ and bisector of angle $ABC$ - centrum of circle $e$ .
5. Find length $GH=4$ .

Answer $4$ .

I know there are official standard solutions for this problem, but I just wanted to point out a 'shortcut' by manipulating the unintended hints of
1) the answer is an integer (also a positive one considering that it's a real distance of a circle's radius), OR
2) the way that it accepted an estimate as long as it lies within allowed range (only if the answer wasn't an integer).
[ Disclaimer : only applicable to this site, NEVER do this on a pen & paper examination ]

The answer should be between the smallest radius when the semicircle lies within the triangle and the greatest radius when the circle touched the middle of chord with length 12 and the bigger circle internally within the major segment.

Let the smallest radius be x.
x / (12 – x) = (5 – x) / x
x² = (12 – x) × (5 – x)
= x² – 17x + 60
x = 60/17 = 3.53

Let the greatest radius be y.
2y = [ 5 + √(5² + 12²) ] / 2
= [ 5 + 13 ] / 2
= 18 / 2
= 9
y = 9/2 = 4.5

My first attempt would be the average of these answers, but it would have shown an "integer only" reminder and the only integer possible is 4.

For a right triangle that circle has radius twice the inradius. In our case it is 2*2=4.

I do not have a proof, just an observation.