Here ABC is a right angle triangle. BC is the Hypotenuse of the triangle.

We know, diameter of circumcircle of a right angle triangle is equal to the Hypotenuse of the triangle also midpoint of Hypotenuse is the center of circle.

$$\therefore$$ $$BC$$ = Diameter of the circle

Here $$B = (0,1)$$ and $$C(3,0)$$

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

$$ = \sqrt {9 + 1} $$

$$ = \sqrt {10} $$

$$\therefore$$ Radius of circumcircle $$(R) = {{\sqrt {10} } \over 2}$$

$$\therefore$$ Center of circle $$(M) = \left( {{{3 + 0} \over 2},\,{{0 + 1} \over 2}} \right) = \left( {{3 \over 2},\,{1 \over 2}} \right)$$

Center of circle which touches line AB and AC $$ = (r,r)$$

Now distance between center of two circles,

$$ME = R - r = {{\sqrt {10} } \over 2} - r$$

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

$$ \Rightarrow {r^2} - 3r + {9 \over 4} + {r^2} - r + {1 \over 4} = {{10} \over 4} + {r^2} - \sqrt {10} r$$

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

$$ \Rightarrow r(r - 4 + \sqrt {10} ) = 0$$

$$ \Rightarrow r = 0$$ or $$r = r - \sqrt {10} $$

$$\therefore$$ $$r = 4 - \sqrt {10} $$ [as $$r \ne 0$$]

$$ = 0.837$$

$$ \simeq 0.84$$