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

Jul 1, 2016

circle A enclosed inside circle B

#### Explanation:

What we have to do here is compare the distance (d) between the centres with the sum and difference of the radii.

• If sum of radii > d , then circles overlap

• If sum of radii < d , then no overlap

• If difference of radii > d , then 1 circle inside other

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 the centres (3 ,2) and (1 ,3)

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

d=sqrt((3-2)^2+(1-3)^2)=sqrt(1+4)=sqrt5≈2.236

sum of radii = radius of A + radius of B = 1 + 4 = 5

difference of radii = radius of B - radius of A = 4 - 1 = 3

Since difference of radii > d , then circle A is enclosed inside circle B.
graph{(y^2-4y+x^2-6x+12)(y^2-6y+x^2-2x-6)=0 [-10, 10, -5, 5]}