# 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?

May 20, 2016

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 $C 1$) 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 $d \le R + r$,
AND, at the same time, (condition $C 2$) 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 $d \ge R - r$.

The distance between $A \left(1 , 2\right)$ and $B \left(- 3 , 3\right)$ is
$d = \sqrt{{\left(- 3 - 1\right)}^{2} + {\left(3 - 2\right)}^{2}} = \sqrt{17} \approx 4.1231$

The sum of their two radiuses is
$R + r = 8 + 4 = 12 > d$
(so, condition $C 1$ is true)

The difference between their two radiuses is
$R - r = 8 - 4 = 4 < d$
(so, condition $C 2$ is true)

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