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

Aug 27, 2016

no overlap, ≈ 1.325

#### Explanation:

What we have to do here is compare the distance (d) between the centres of the circles to 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 $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points}$

the 2 points here are (-4 ,-3) and (2 ,-1) the centres of the circles.

let $\left({x}_{1} , {y}_{1}\right) = \left(- 4 , - 3\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(2 , - 1\right)$

d=sqrt((2+4)^2+(-1+3)^2)=sqrt(36+4)=sqrt40≈6.325

sum of radii = radius of A + radius of B = 2 + 3 = 5

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

min. distance between them = d - sum of radii

$= 6.325 - 5 = 1.325$
graph{(y^2+6y+x^2+8x+21)(y^2+2y+x^2-4x-4)=0 [-10, 10, -5, 5]}