The radius of the circle is 12.\frac 12.21​. This makes the total possible region for the circle's center to be in a square of length 8−12−12=7.8 - \frac 12 - \frac 12 = 7.8−21​−21​=7. Shown below is the region in which the circle can be.

Now, the circle's center must be inside the black region or within 12\frac 1221​ of the black region.

For each of the corner, we can take a right triangle for this region. Now, we can find the altitude to the hypotenuse. This would be useful to find the as the area of a right triangle is equal to the altitude times the hypotenuse divided by 2,2,2, and the hypotenuse is twice the altitude, making the area equal to the altitude squared.

The altitude of this region is equal to the altitude of the portion of the right triangle in the black region and the blue region plus 12.\frac 12.21​. When finding the portion of the black right triangle in the blue region, we get a right triangle with length 2−12−12=1.2-\frac 12 - \frac 12 = 1.2−21​−21​=1. This has an altitude of 22.\frac{\sqrt 2}2 .22​​. The total altitude therefore is 2+12.\frac{\sqrt 2 + 1}2.22​+1​. Thus, the answer is (2+12)2=3+224.(\dfrac{\sqrt 2 + 1}2)^2 = \dfrac{3+2\sqrt 2}4.(22​+1​)2=43+22​​. There are 444 corners, so the combined area of the right triangles is 3+22.3+ 2\sqrt 2.3+22​.

For the center square, we find the area within it, the area that is within 12\frac 1221​ of the edge, and within 12\frac 1221​ of the corners. The area in the square is (22)2=8.(2\sqrt 2)^2 = 8.(22​)2=8. The area within the 12\frac 1221​ of the edges is 12⋅22=2\frac 12 \cdot 2\sqrt 2 = \sqrt 2 21​⋅22​=2​ since its a rectangle of with lengths 12\frac 12 21​ and 22.2\sqrt 2.22​. The total area from this is 42.4\sqrt 2.42​. The area within 12\frac 1221​ of the points are 4 quarter circles of radius 12,\frac 12,21​, so their combined area is π4.\frac \pi 4 .4π​. The total area is 3+22+9+42+π43+ 2\sqrt 2 + 9 + 4\sqrt 2 + \frac \pi 43+22​+9+42​+4π​ =8+62+π4.=8 + 6\sqrt 2 + \frac \pi 4.=8+62​+4π​.

The probability is 11+62+π449\dfrac{11 + 6\sqrt 2 + \frac \pi 4} {49}4911+62​+4π​​ =44+242+π196.=\dfrac{44 + 24\sqrt 2 + \pi } {196}.=19644+242​+π​. This makes the answer 44+24=68.44+24=68.44+24=68.

Thus, the answer is C.