See this earlier MO question. The union of n disks (represented in terms of its boundary arcs) can be constructed in $O(n\log n)$ time. So construct the union of the $n$ disks you are given, and separately construct the union of the $n+1$ disks including the one you want covered. The given disk is covered if and only if the two unions are the same.

See this earlier MO question. The union of n disks (represented in terms of its boundary arcs) can be constructed in $O(n\log n)$ time. So construct the union of the $n$ disks you are given, and separately construct the union of the $n+1$ disks including the one you want covered. The given disk is covered if and only if the two unions are the same.

See this earlier MO question. The union of n disks (represented in terms of its boundary arcs) can be constructed in $O(n\log n)$ time. So construct the union of the $n$ disks you are given, separately construct the union of the $n+1$ disks including the one you want covered. The given disk is covered iff and only if the two unions are the same.

See this earlier MO question. The union of n disks (represented in terms of its boundary arcs) can be constructed in $O(n\log n)$ time. So construct the union of the $n$ disks you are given, and separately construct the union of the $n+1$ disks including the one you want covered. The given disk is covered if and only if the two unions are the same.