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 $\qquad$

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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 $F_{1}$ about the line $l$ be $F_{1}^{\prime}$.
Then $F_{1}^{\prime} 、 F_{2} 、 M$ are collinear, and
$\left|F_{1}^{\prime} F_{2}\right|=\left|F_{1}^{\prime} M\right|+\left|M F_{2}\right|$
$=\left|F_{2} M\right|+\left|M F_{1}\right|=2$.
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 $F_{2} 、 F_{1}$ are on the same side of the line $l$, the trajectory equation of point $F_{2}$ is
$$
(x+1)^{2}+(y-3)^{2}=4(-1<x<1,1<y<3) .
$$