4. The solution set $(x, y)$ that satisfies the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}-14 x-10 y+58=0, \\
\sqrt{x^{2}+y^{2}-16 x-12 y+100}+\sqrt{x^{2}+y^{2}+4 x-20 y+104}=2 \sqrt{29}
\end{array}\right.
$$

is ( ).
(A) $\left(\frac{217+5 \sqrt{415}}{29}, \frac{180-2 \sqrt{415}}{29}\right)$
(B) $\left(\frac{180-2 \sqrt{415}}{29}, \frac{217+5 \sqrt{415}}{29}\right)$
(C) $\left(\frac{217-5 \sqrt{415}}{29}, \frac{180+2 \sqrt{415}}{29}\right)$
(D) $\left(\frac{180+2 \sqrt{415}}{29}, \frac{217-5 \sqrt{415}}{29}\right)$
4.C.

The system of equations becomes
$$
\left\{\begin{array}{l}
(x-7)^{2}+(y-5)^{2}=4^{2}, \\
\sqrt{(x-8)^{2}+(y-6)^{2}}+\sqrt{(x+2)^{2}+(y-10)^{2}} \\
=\sqrt{10^{2}+4^{2}} .
\end{array}\right.
$$

From (1), we know that the solution $(x, y)$ lies on a circle with center $(7,5)$ and radius 4; from (2), we know that the solution $(x, y)$ lies on the line segment connecting the points $(8,6)$ and $(-2,10)$. The equation of the line passing through the points $(8,6)$ and $(-2,10)$ is
$2 x+5 y=46$.
(3) combined with (1) gives the two intersection points of the line and the circle:
$$
\begin{array}{l}
\left(\frac{217-5 \sqrt{415}}{29}, \frac{180+2 \sqrt{415}}{29}\right) \\
\text { and }\left(\frac{217+5 \sqrt{415}}{29}, \frac{180-2 \sqrt{415}}{29}\right) .
\end{array}
$$

Since $\frac{217+5 \sqrt{415}}{29}>10>8$, the second intersection point lies on the extension of the line segment with endpoints $(8,6)$ and $(-2,10)$, and should be excluded. Only the first intersection point $\left(\frac{217-5 \sqrt{415}}{29}, \frac{180+2 \sqrt{415}}{29}\right)$ meets the requirements (this can also be determined by drawing a rough sketch).