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

Jun 16, 2016

no overlap , ≈ 8.18

#### Explanation:

What we have to do here is 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 $\left({x}_{1} , {y}_{1}\right) = \left(2 , 8\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(- 8 , 3\right)$

d=sqrt((-8-2)^2+(3-8)^2)=sqrt(100+25)≈11.18

radius of A + radius of B = 2 + 1 = 3

Since sum of radii < d , then no overlap

smallest distance = d - sum of radii = 11.18 - 3 = 8.18
graph{(y^2-16y+x^2-4x+64)(y^2-6y+x^2+16x+72)=0 [-20, 20, -10, 10]}