23. As shown in Figure 3, a white square with a side length of 8 has four black isosceles right triangles with leg lengths of 2 placed at its four corners. A black square with a side length of \(2 \sqrt{2}\) is placed at the center of the square. A circular coin with a diameter of 1 is randomly placed at some position inside the square. The probability that the coin covers a black area of the square can be expressed as \(\frac{a+b \sqrt{2}+\pi}{196}\left(a, b \in \mathbf{Z}_{+}\right)\). Then the value of \(a+b\) is ( ).
(A) 64
(B) 66
(C) 68
(D) 70
(E) 72
23. C.

From the problem, we have
$$
\begin{array}{l} 
P=\frac{4 \times \frac{1}{2}\left(\frac{\sqrt{2}}{2}+1\right)^{2}+4 \sqrt{2}+8+\frac{\pi}{4}}{7^{2}} \\
=\frac{44+24 \sqrt{2}+\pi}{196} .
\end{array}
$$

Thus, $a=44, b=24$.
Therefore, $a+b=68$.