Sky Region

Just for sake of variety .....

Let $A(0,0), B(0,1), C(1,1), D(1,0)$ be the corners of the square, $P(\frac{1}{4}, \frac{1}{4})$ be the center of the circle, $Q$ be the lower left vertex of the triangle and $T$ be the point of tangency of the circle and the triangle.

The area of the triangle will then be $\frac{1}{2}\tan(\angle QCD),$ (since the side length of the square is $1.$ )

Now $\angle QCD = \dfrac{\pi}{2} - \angle BCP - \angle PCT = \dfrac{\pi}{2} - \dfrac{\pi}{4} - \sin^{-1}\left(\dfrac{|PT|}{PC|}\right) =$

$\dfrac{\pi}{4} - \sin^{-1}\left(\dfrac{\frac{1}{4}}{\frac{3}{4}\sqrt{2}}\right) = \dfrac{\pi}{4} - \sin^{-1}\left(\dfrac{\sqrt{2}}{6}\right).$

So $\tan(\angle QCD) = \dfrac{1 - \tan\left(\sin^{-1}(\frac{\sqrt{2}}{6})\right)}{1 + \tan\left(\sin^{-1}(\frac{\sqrt{2}}{6})\right)} =$

$\dfrac{1 - \frac{\sqrt{17}}{17}}{1 + \frac{\sqrt{17}}{17}} = \dfrac{\sqrt{17} - 1}{\sqrt{17} + 1} = \dfrac{9 - \sqrt{17}}{8}.$

The area of the triangle is then $\dfrac{9 - \sqrt{17}}{16},$ and so

$a + b - c = 9 + 17 - 16 = \boxed{10}.$

Let $A$ , $B$ , $C$ , $D$ be the corners of the square and $P$ be the center of the circle. $E$ is the lower left vertex of the triangle, $F$ is the point of tangency of the circle and $CE$ , $G$ is the point of tangency of the circle and $AB$ .

$CE$ is tangent to the circle, so it's perpendicular to the radius $PF$ .

$\angle PFC = 90^\circ$

The radius of the circle is $\frac 1 4$ of the side of the square, so $PC$ is $(1 - \frac 1 4)$ of the diagonal of the square.

$\overline {PC} = \frac {3} {4} \times \sqrt {2} = \frac {3\sqrt{2}} {4}$

$\overline {PF} = \frac {1} {4}$

In the right triangle $PFC$ ,

$\overline {CF}^2 = \overline {PC}^2 - \overline {PF}^2$

by the Pythagorean theorem.

So $\overline {CF} = \sqrt {\overline {PC}^2 - \overline {PF}^2} = \sqrt {{\frac {3\sqrt{2}} {4}}^2 - {\frac {1} {4}}^2} = \sqrt {\frac {18} {16} - \frac {1} {16}} = \sqrt {\frac {17} {16}} = \frac {\sqrt {17}} {4}$

Both $EF$ and $GE$ are tangent to the circle and start from the same point $E$ . So $\overline {EF} = \overline {GE}$

$AG$ is equal to the radius of the circle, so $\overline {AG} = \frac {1} {4}$

$\overline {AB} = \overline {AG} + \overline {GE} + \overline {EB} = 1$

So $\overline {GE} + \overline {EB} = \overline {AB} - \overline {AG} = 1 - \frac {1} {4} = \frac {3} {4}$

$\overline {GE} = \frac {3} {4} - \overline {EB}$

$\overline {GE} = \overline {FE} = \overline {CE} - \overline {CF} = \overline {CE} - \frac {\sqrt {17}} {4}$

And then $\frac {3} {4} - \overline {EB} = \overline {CE} - \frac {\sqrt {17}} {4}$

$\overline {CE} = \frac {3+\sqrt{17}} {4} - \overline {EB}$

In the right triangle $CBE$ ,

$\overline {EB}^2 = \overline {CE}^2 - \overline {BC}^2 = \overline {CE}^2 - 1$

by the Pythagorean theorem.

$\overline {EB}^2 = ( \frac {3+\sqrt{17}} {4} - \overline {EB} )^2 - 1 = \frac {26 + 6 \sqrt{17} } {16} - \frac {3+\sqrt{17}} {2} \overline {EB} + \overline {EB}^2 -1$

$\frac {3+\sqrt{17}} {2} \overline {EB} = \frac {26 + 6 \sqrt{17} } {16} - 1 = \frac {26 + 6 \sqrt{17} - 16} {16} = \frac {10 + 6 \sqrt{17} } {16} = \frac {5 + 3 \sqrt{17}} {8}$

$\overline {EB} = \frac {5 + 3 \sqrt{17}} {8} \times \frac {2} {3+\sqrt{17}} = \frac {9 - \sqrt{17}} {8}$

Now, the area of the triangle $CBE$ is

$A = {\frac {1} {2}} \times {\overline {EB}} \times {\overline {BC}} = {\frac {1} {2}} \times {\overline {EB}} = {\frac {1} {2} } \times {\frac {9 - \sqrt{17} } {8}} = \frac {9 - \sqrt{17} } {16}$

The answer is $9 + 17 - 16 = \boxed{10}$

My approach was somewhat peasant way by using analytic geometry.

The sought area will be 1/2 (side)^2 times m, being m the slope of the vertex of sky region farthest from circle.

Procedure

1) First without loosing generality, we can assume side of square =4. and locate the circle tangent to it at the upper right corner so the resulting equations would be more handy, we are looking just for m

2) We take an straight line passing through the origin with an slope m.

3) Writing both equation and without solving completely we need just find the value of m that make the discriminant equal to 0 by doing so we get the values of m for the double roots that force precisely to be tangent,

Following the above we get for the for circle : (x-3)^2 + (y-3)^2 =1

for the straight line y =mx.

By substitution of y in circle equation we obtain (1+m^2)x^2 -6(1+m) +17 = 0

Forcing the discriminant equal to zero we get the equation after simplifying

4m^2- 9m +4 =0

which solution render the two possiblevalues of m tangents to the circle.

(9-17^.5)/8 and (9+17^.5)/8.

Therefore the sought area must be = (9-17^.5)/16