# Circle A has a center at (3 ,7 ) and a radius of 1 . Circle B has a center at (-3 ,-2 ) and a radius of 2 . Do the circles overlap? If not, what is the smallest distance between them?

Sep 21, 2017

Shortest distance $\approx 7.187$

#### Explanation:

When given the coordinates, the shortest distance between two disjoint circles (circles wihich don't overlap), is the distance between the two centrepoints minus the radii of each circle. The equation looks like this:

$D = \left(\sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}\right) - {r}_{1} - {r}_{2}$, where
$D$ is the distance, ${r}_{1}$ isthe first circle's radius, ${r}_{2}$ is the second circles radius, $\left({x}_{1} , {y}_{1}\right)$ is the first circles centrepoint, and $\left({x}_{2} , {y}_{2}\right)$ is the second circles centrepoint. This equation is modelled off of the Pythagoras Theorem, as shown below: https://stackoverflow.com/questions/10666116/most-efficient-way-to-find-distance-between-two-circles-in-java

From this picture, you can see why the shortest distance between two disjoint circles is their centrepoint distance minus their two radius, shown in the picture as $h$.

We know that ${r}_{1} = 1$, $\left({x}_{1} , {y}_{1}\right) = \left(3 , 7\right)$, ${r}_{2} = 2$ and $\left({x}_{2} , {y}_{2}\right) = \left(- 3 , - 2\right)$. Plug these values into the equation, and we get
$D = \left(\sqrt{{\left(3 - - 3\right)}^{2} + {\left(7 - - 2\right)}^{2}}\right) - 1 - 2$
$D = \left(\sqrt{{6}^{2} + {9}^{2}}\right) - 3$
$D = \left(\sqrt{117}\right) - 3$
$D = \left(3 \sqrt{13}\right) - 3$
$D \approx 7.187$, or the smallest distance between the two circles is roughly 7.187.

I hope that helped!