## Elementary Geometry for College Students (7th Edition)

$\sqrt{5}$
We know that the center of the circumscribed circle is the midpoint of the line BA. Assuming that point C of the triangle is the origin, this means that the midpoint is (4,3). We now consider the midpoint of the inscribed circle, which is given by: $r =\frac{ab}{a+b+c} \\ r = \frac{8\times 6}{8+6+10} = 2$ Thus, we use the distance formula to find the distance: $d = \sqrt{(4-2)^2 + (3-2)^2} = \sqrt{5}$