5. Given real numbers $x, y$ satisfy $x^{2}+y^{2}=1$. Then the sum of the maximum and minimum values of $(x-3)^{2}+(y+4)^{2}$ is ( ).
(A) 48
(B) 49
(C) 50
(D) 52
(A) $2 \sqrt{2}$
(B) $1.5 \sqrt{2}$
(C) $3 \sqrt{3}$
(D) $2 \sqrt{3}$
5. D.

Let $(x-3)^{2}+(y+4)^{2}=t$. Then
$$
\begin{array}{l}
t \geqslant 0, \\
t=x^{2}+y^{2}-6 x+8 y+25=-6 x+8 y+26 \\
\Rightarrow y=\frac{3}{4} x+\frac{t-26}{8} .
\end{array}
$$

Substitute into $x^{2}+y^{2}=1$, simplify and organize to get
$$
\begin{array}{l}
100 x^{2}+12(t-26) x+(t-26)^{2}-64=0 \\
\Rightarrow \Delta=(12(t-26))^{2}-4 \times 100\left((t-26)^{2}-64\right) \\
\quad \geqslant 0 \\
\Rightarrow 16 \leqslant t \leqslant 36 .
\end{array}
$$

Therefore, the maximum and minimum values of $t$ are $36$ and $16$, respectively, and the answer is 52.