# Circle A has a center at (-1 ,3 ) and a radius of 5 . Circle B has a center at (3 ,-2 ) and a radius of 1 . Do the circles overlap? If not what is the smallest distance between them?

Dec 20, 2016

"no overlap, smallest distance" ≈0.403

#### Explanation:

What we have to do here is $\textcolor{b l u e}{\text{compare}}$ the distance (d ) between the centres of the circles to the $\textcolor{b l u e}{\text{sum of the radii}}$

• If sum of radii > d , then circles overlap

• If sum of radii < d , then no overlap

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

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

The 2 points here are (-1 ,3) and (3 ,-2)

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

d=sqrt((3+1)^2+(-2-3)^2)=sqrt(16+25)=sqrt41≈6.403

Sum of radii = radius of A + radius of B = 5 + 1 = 6

Since sum of radii < d , then no overlap

smallest distance = d - sum of radii

$= 6.403 - 6 = 0.403$
graph{(y^2-6y+x^2+2x-15)(y^2+4y+x^2-6x+12)=0 [-20, 20, -10, 10]}