# Circle A has a center at (2 ,-7 ) and a radius of 2 . Circle B has a center at (-3 ,6 ) and a radius of 1 . Do the circles overlap? If not, what is the smallest distance between them?

May 27, 2016

no overlap , ≈ 10.928

#### 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 $\left({x}_{1} , {y}_{1}\right) = \left(2 , - 7\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(- 3 , 6\right)$

d=sqrt((-3-2)^2+(6+7)^2)=sqrt(25+169)≈13.928

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

Since sum of radii < d , there is no overlap.

smallest distance between them = 13.928 - 3 =10.928
graph{(y^2+14y+x^2-4x+49)(y^2-12y+x^2+6x+44)=0 [-20, 20, -10, 10]}