# Circle A has a center at (3 ,7 ) and a radius of 4 . Circle B has a center at (4 ,-2 ) and a radius of 6 . Do the circles overlap? If not, what is the smallest distance between them?

Aug 15, 2016

circles overlap.

#### 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 (3 ,7) and (4 ,-2) the centres of the circles.

let $\left({x}_{1} , {y}_{1}\right) = \left(3 , 7\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(4 , - 2\right)$

d=sqrt((4-3)^2+(-2-7)^2)=sqrt(1+81)=sqrt82≈9.055

sum of radii = radius of A + radius of B = 4 + 6 = 10

Since sum of radii > d , then circles overlap
graph{(y^2-14y+x^2-6x+42)(y^2+4y+x^2-8x-16)=0 [-40, 40, -20, 20]}