Maximum ratio

Let the other leg be of length $x$ . Then the ratio of the radii of the incircle ( $r$ ) and the Circumcircle ( $R$ ) is $\eta=\dfrac{r}{R}=\dfrac{2x}{x^2+1+(x+1)\sqrt {x^2+1}}$ .

This will be maximum when it's first derivative with respect to $x$ is zero, that is,

$x^3-1+(x^2-1)\sqrt {x^2+1}=0\implies x=1$

The minimum value of the ratio of the area of the two circles is $\eta^2=(\sqrt 2 -1)^2=3-2\sqrt 2$

Hence, $a=3,b=c=2,$ and $a+b+c=3+2+2=\boxed 7$ .

Let the length of the other leg be x. Then, the hypotenuse of the triangle (by the Pythagorean Theorem) is sqrt(1+x^2). It is known that the hypotenuse of the triangle is also a diameter of the circumscribed circle. Hence, the area of the circumscribed circle is Pi [(1/2) (sqrt(1+x^2)]^2. By dropping perpendicular lines from the center of the inscribed circle to the points of tangency of the triangle and the inscribed circle and comparing areas, we can show that the radius of the inscribed circle is given by R = x*(1+x+sqrt(1+x^2))^-1. Hence, the area of the inscribed circle is given by the quantity:

Area inscribed circle = Pi ((x (1+x+sqrt(1+x^2))^-1)^2.

Dividing the two areas above and simplifying yields the following function to be minimized:

f(x) = (4 x^2) / ((1+x^2) (1+x+sqrt(x^2+1))^2)

Standard (but messy) calculations (taking the derivative, setting equal to zero, etc.) show that f is maximized at x = 1. The maximum value is:

f(1) = 3-2*sqrt(2), so that a = 3, b = 2, and c = 2. Hence, a+b+c = 7.