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

Feb 8, 2016

Smallest distance between the two circles is $\sqrt{61} - 5 \approx 2.81$.

#### Explanation:

Let's compute the distance between the two centers of the circle.

This can be done with Pythagorean theorem:

${d}^{2} = {d}_{x}^{2} + {d}_{y}^{2}$

where $d$ is the distance between the two points, ${d}_{x}$ is the distance between the $x$ values and ${d}_{y}$ is the distance between the $y$ values of the points.

In your case, you have

${d}^{2} = {\left(1 - \left(- 4\right)\right)}^{2} + {\left(- 2 - \left(- 8\right)\right)}^{2} = 25 + 36 = 61$

$\implies d = \sqrt{61} \approx 7.81$

However, $7.81$ is the distance between the centers and not the distance between the outer points of the circle.
To compute the distance between the circles, we also need to take the radius ${r}_{1} = 3$ and ${r}_{2} = 3$ into consideration:

$\text{smallest distance} = d - {r}_{1} - {r}_{2} = d - 3 - 2 = \sqrt{61} - 5 \approx 2.81$

Thus, the smallest distance between these two circles is $\approx 2.81$.

I will insert a graph of the two circles with the line that passes through both centers.
The smallest distance is the length of this line between the two circles.

graph{((x-1)^2 + (y+2)^2 - 9)((x+4)^2 + (y+8)^2 - 4)(y - (6/5 x - 16/5)) = 0 [-13.92, 11.4, -10.73, 1.93]}