# Circle A has a center at (8 ,3 ) and a radius of 1 . Circle B has a center at (4 ,4 ) and a radius of 2 . Do the circles overlap? If not what is the smallest distance between them?

Jul 28, 2016

no overlap, ≈ 1.123

#### Explanation:

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

the 2 points here are (8 ,3) and (4 ,4) the centres of the circles.

let $\left({x}_{1} , {y}_{1}\right) = \left(8 , 3\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(4 , 4\right)$

d=sqrt((4-8)^2+(4-3)^2)=sqrt(16+1)=sqrt17≈4.123

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

Since sum of radii < d , then no overlap

smallest distance = d - sum of radii = 4.123 - 3 = 1.123
graph{(y^2-6y+x^2-16x+72)(y^2-8y+x^2-8x+28)=0 [-10, 10, -5, 5]}