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

May 31, 2016

no overlap , ≈ 4.662

#### Explanation:

What we have to do here is to compare the distance (d ) between the centres with the sum of the radii.

• If sum of radii > d , then circles overlap

• If sum of radii < d , then no overlap

To calculate d , use the $\textcolor{b l u e}{\text{distance formula}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 points}$

let $\left({x}_{1} , {y}_{1}\right) = \left(8 , 4\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(- 2 , - 2\right)$

d=sqrt((-2-8)^2+(-2-4)^2)=sqrt136≈11.662

radius of A + radius of B = 3 + 4 = 7

Since sum of radii < d , then no overlap

and smallest distance between them = 11.662 - 7 = 4.662
graph{(y^2-8y+x^2-16x+71)(y^2+4y+x^2+4x-8)=0 [-20, 20, -10, 10]}