How to represent circles in x-y coordinates
This answer considers two cases:


the overlapping relation between two disks, which is a very simple problem.
the ovelapping or covering of a disk by a set of other disks, which is somewhat harder in general.


Case of two disks

It is indeed a good idea to use center and radius to represent your
circles. However I think you are not thinking of circles, which are
closed planar lines formed of all points at a given distance $r$,
called radius, of a point $c$ called the center. The planar surface
enclosed by a circle, which also includes all points at a lesser
distance fron $c$ is called a
disk (also
spelled disc).

Regarding the various overlap situations for two disks, you have the
following test, assuming the two circles have radius $r_1$ and $r_2$
(assuming witout loss of generality that $r_1\geq r_2$), with their
centers at distance $d$ (computable from the centers coordinates,
thanks to Pythagoras):


$r_1+r_2&lt;d$ : disks are disjoint, no overlap.
$r_1+r_2=d$ : disks are tangent externally, a single point of overlap.
$r1-r2&lt;d&lt;r_1+r_2$ : disks are partially overlapping.
$r_1-r_2=d$ : disks are tangent internally, total overlap of disk 2 by disk 1.
$d&lt;r_1-r_2$ : disk 1 overlap totally disk 2.


There can be exact overlap of each disk by the other only when
$r_1=r_2 \wedge d=0$

Regarding the case of several disk, it is not clear whether you want
to see whether one distinguished disk is overlapped totally, partially
or not at all by the others, or whether you want to check that for
each disk with everyone of the others, or possibly something else. You should make that more precise.

Case of several disks

This is only a rough sketch, hopefully correct. Working out all details is a lot more work than can be included in an answer.

As suggested in the question, we can represent a disk $D$ by a
triple $(x,y,r)$ which gives the coordinates of the center, and the radius.

Now if you have a disk $D_0=(x_0,y_0,r_0)$ and a set of disks
$L=\{D_i\mid D_i=(x_i,y_i,r_i),\; i\in[1,n]\}$, your question, as
made more precise in a comment, is
how to check whether the disks of set $L$ together overlap partially,
totally or not at all the disk $D_0$.

First you want to make a list of the disks in $L$ that actually
overlap $D_0$. For that you can simply apply the above test to $D_0$ and
$D_i$ for every $D_i$ in $L$. You get a set $I\subseteq[1,n]$ of indices
such that D overlaps every $D_i$ for $i\in I$.

If this set $I$ is not empty, you know that $D_0$ is overlapped by $L$.

The question remains of a total overlap of $D_0$ by $L$.

For this, you create a new set $J\subseteq I$ by removing indices of
disks $D_i$ that are only externally tangent to $D_0$, overlapping it on
only one point.

If $J$ is empty, you had only tangential overlapping in a finite
number of point, but the set $L$ does not cover (fully overlap) the disk $D_0$.

If $J$ is not empty, then $L$ completely overlap $D_0$ iff it does it
with the disks $D_i$ such that $i\in J$. Tangential overlapping on a
single point cannot contribute usefully to overlapping a surface.

Now, all disks in $M=\{D_i\mid i\in J\}$ overlap $D_0$ on a fragment
of its surface.

If one disk in $M$ completely overlaps $D_0$, then we have an answer
of complete overlap. We can now assume it is not the case.

Then the problem is to find an orderly strategy to check whether the
disks in $M$ completely cover $D_0$. We will now use disks from $M$
one by one, removing them from M, to cover $D_0$. The strategy is in
choosing them.

We chose first a disk in M (which we remove from M), such that it is not
internally tangent to $D_0$, but intersect the edge circle of $D_0$. There
must be at least two such disks, else $D_0$ cannot be covered, as its
perimeter circle will nor be covered except for a finite number of
points.

We compute the two points where the two circles intersect. They define
two arcs, one on each circle, that delimit a surface yet to be covered
by the other circles. With respect to this surface, the arc belonging
to $D_0$ is convex, while the other is concave.

For simplicity and intuition, we call these intersections the angles
of the remaining surface. We call sides or edges the arcs between 2
angles that delimit the remaining surface to be covered.

We then chose one of the two angles, and look in M for another circle
covering this intersection. This new circle removes at least this
angle, and adds usually two new angles. We discuss this further below.

One remark is that the arcs coming from $D_0$ are always convex, while
the others are always concave. This is useful to visualize what can occur.

Then we repeat the same step, until there are no angles left, in which
case we have a covering of $D_0$, which is tested first, or until
there are no disk left in M that can cover a chosen angle, in which
case the overlap is incomplete.

To understand this we have to look at intermediate steps in more
details. During execution of the algorithm, we may have actually
created an uncovered curvy polygon with many angles. When we chose a
new disk $D_i$ to cover a chosen angle $A$, we may actually cover
several angles of the polygon, and the edges in between, so that we
actually reduce the number of edges.

So after choosing the disk $D_i$, we have to find the first edge on
both sides of the angle $A$ (not necessarily adjacent to $A$) that are
intersected by the circle bounding $D_i$. These two intersections, one
on each side, define two new angles and a new edge provided $D_i$, and
may replace several edges and angles. Hence the number of sides of the
polygon may be reduced rather than increased. It may be also that all
sides are covered, in wich case we have covered the whole disk.

There may be cases when the new disk will cut the curvy polygon into two
polygons, that will both have to be covered. To check for that, the
circle bounding the disk has to be checked for intersection with all
edges of the curvy polygon(s).

All this implies of course to keep an up-to-date description of the
curvy polygon and the relations between angles, edges and disks.
To keep correspondence between the relative positions of angles, if
they are listed counter-clockwise on the curvy polygon(s), then they
should be counter-clockwise on the circle of the disk $D_0$ and
clockwise on the disks from $M$.

There are quite a few details to be worked out, which I would not try
without at least checking an implementation. But I believe this basic
idea can be implemented.

From this analysis, I think the time complexity is $O(n^2)$, since
every steps considers a new disk, and has to look at intersections
potentially with all disks that have already been considered.

The extension to spheres is left as an exercise.