4. Given that one focus of the ellipse $C$ is $F_{1}(1,1)$, the length of the major axis is 2, and the line $l: y=x+2$ is tangent to the ellipse. Then the equation of the trajectory of the other focus $F_{2}$ of the ellipse $C$ is $\qquad$ .
4. $(x+1)^{2}+(y-3)^{2}=4(-1<x<1$, $1<y<3)$.

Let the point of tangency between the ellipse $C$ and the line $l$ be $M$, and the symmetric point of $F_{1}$ with respect to the line $l$ be $F_{1}^{\prime}$. 
Then $F_{1}^{\prime}, F_{2}, M$ are collinear, and
$$
\begin{array}{l}
\left|F_{1}^{\prime} F_{2}\right|=\left|F_{1}^{\prime} M\right|+\left|M F_{2}\right| \\
=\left|F_{2} M\right|+\left|M F_{1}\right|=2 .
\end{array}
$$

Thus, the point $F_{2}$ lies on the arc of a circle with $F_{1}^{\prime}(-1,3)$ as the center and 2 as the radius.

Since the points $F_{2}$ and $F_{1}$ are on the same side of the line $l$, the trajectory equation of point $F_{2}$ is
$$
(x+1)^{2}+(y-3)^{2}=4(-1<x<1,1<y<3) \text {. }
$$