# Circle A has a center at (4 ,-1 ) and a radius of 3 . Circle B has a center at (-3 ,6 ) and a radius of 2 . Do the circles overlap? If not, what is the smallest distance between them?

Sep 6, 2016

no overlap, min. distance ≈ 4.899

#### 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 ,-1) and (-3 ,6) the centres of the circles.

let $\left({x}_{1} , {y}_{1}\right) = \left(4 , - 1\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(- 3 , 6\right)$

d=sqrt((-3-4)^2+(6+1)^2)=sqrt(49+49)=sqrt98≈9.899

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

Since sum of radii < d , then no overlap

smallest distance = d - sum of radii = 9.899 - 5 = 4.899
graph{(y^2+2y+x^2-8x+8)(y^2-12y+x^2+6x+41)=0 [-20, 20, -10, 10]}