Dürer explains art history to his students. The following gothic window is examined.
Where the center of the arc of $BC$ is $A$, and similarly the center of the arc of $AC$ is $B$. 
The question is how much is the radius of the circle (radius marked $r$ in the figure).[img]https://cdn.artofproblemsolving.com/attachments/5/c/28e5ee47005bfde7f925908b519099d5e28d91.png[/img]
1. Define the center of the unknown circle as \( P \). We need to find the radius \( r \) of the circle centered at \( P \).

2. Given that the center of the arc \( BC \) is \( A \) and the center of the arc \( AC \) is \( B \), we can infer that \( A \) and \( B \) are equidistant from \( P \).

3. Let \( O \) be the center of the circle passing through \( A \) and \( B \). The radius of this circle is \( r \).

4. We know that \( OP^2 = r^2 + 2r \). This is derived from the Pythagorean theorem in the right triangle formed by \( O \), \( P \), and the point where the radius \( r \) meets the circle perpendicularly.

5. Therefore, \( AP^2 = r^2 + 2r + 4 \). This is because \( A \) is 2 units away from \( O \) (the radius of the smaller circle), and we add this distance squared to the previous equation.

6. Let \( Q \) be the intersection of \( AP \) with the arc \( BC \). Note that \( AQ = 4 \).

7. We also have that \( AQ = r + \sqrt{r^2 + 2r + 4} = 4 \).

8. Solving the equation \( r + \sqrt{r^2 + 2r + 4} = 4 \):

   \[
   \sqrt{r^2 + 2r + 4} = 4 - r
   \]

9. Squaring both sides:

   \[
   r^2 + 2r + 4 = (4 - r)^2
   \]

10. Expanding the right-hand side:

    \[
    r^2 + 2r + 4 = 16 - 8r + r^2
    \]

11. Simplifying the equation:

    \[
    2r + 4 = 16 - 8r
    \]

12. Combining like terms:

    \[
    10r = 12
    \]

13. Solving for \( r \):

    \[
    r = \frac{12}{10} = \frac{6}{5}
    \]

The final answer is \(\boxed{\frac{6}{5}}\).