Question

Circle A has a center at `(1 ,2 )` and a radius of `8`. Circle B has a center at `(-3 ,3 )` and a radius of `4`. Do the circles overlap? If not, what is the smallest distance between them?

Answer 1

Yes, circles overlap.
Hence, the distance between them is equals to zero.

Explanation:

Let `d` be the distance between the centers of two circles of radiuses `R` (bigger) and `r` (smaller).

Non-intersecting circles can be either completely outside of each other or a smaller circle can be completely inside of a bigger one.

In the former case the distance between centers is too big, bigger than the sum of their radiuses.

`d > R+r`

In the latter case the distance between centers is too small, smaller than the difference between a bigger and a smaller radiuses.

`R > d+r` or, equivalently, `d < R-r`

Therefore, to intersect, two circles must have (condition `C1`) the distance between their centers to be not greater than the sum of their two radiuses (otherwise they lie outside each other without intersection), that is `d <= R+r`,
AND, at the same time, (condition `C2`) that same distance between centers must be not less than the difference between a larger and a smaller circles (otherwise, a smaller circle would be completely inside the bigger one without intersection), that is `d>=R-r`.

The distance between `A(1,2)` and `B(-3,3)` is
`d = sqrt((-3-1)^2+(3-2)^2)=sqrt(17)~~4.1231`

The sum of their two radiuses is
`R+r=8+4=12 > d`
(so, condition `C1` is true)

The difference between their two radiuses is
`R-r=8-4=4 < d`
(so, condition `C2` is true)

So, intersection does take place.
Hence, the distance between the circles is zero.