4. Given an ellipse centered at the origin, with foci on the $x$-axis and a focal distance of 4, which is tangent to the line $x+y=9$ at point $P$. Then, the coordinates of point $P$ are ( ).
(A) $\left(\frac{35}{6}, \frac{19}{6}\right)$
(B) $\left(\frac{21}{9}, \frac{60}{9}\right)$
(C) $\left(\frac{85}{18}, \frac{77}{18}\right)$
(D) $\left(\frac{115}{21}, \frac{74}{21}\right)$
4.C.

As shown in Figure 2, let $A$ and $B$ be the foci of the ellipse, and $P$ be any point on the line $x + y = 9$. Then, $P$ is outside the ellipse or coincides with point $P$. Suppose $P'A$ intersects the ellipse at point $Q$, then
$$
P'A + P'B = P'B + PQ + QA \geqslant QB + QA = PA + PB.
$$

Thus, point $P$ is the point on the line $x + y = 9$ with the minimum sum of distances to the two foci of the ellipse.

It is easy to know that $A(-2,0)$, $B(2,0)$, and the symmetric point of point $B$ with respect to the line $x + y = 9$ is $M(9,7)$. Connecting $AM$ intersects the line $x + y = 9$ at point $N\left(\frac{85}{18}, \frac{77}{18}\right)$. For any point $P'$ on the line $x + y = 9$, we have
$$
\begin{array}{l}
P'A + P'B = P'A + P'M \\
\geqslant AM = NA + NM = NA + NB.
\end{array}
$$

Therefore, point $N$ coincides with $P$.