# Circle A has a center at (1 ,4 ) and a radius of 5 . Circle B has a center at (9 ,3 ) and a radius of 1 . Do the circles overlap? If not what is the smallest distance between them?

May 3, 2016

There is a minimum distance of $\left(\sqrt{65} - 6\right) \approx 2.06$ units between the two (non-overlapping) circles.

#### Explanation:

The distance between the two centers is
$\textcolor{w h i t e}{\text{XXX}} d = \sqrt{{\left(9 - 1\right)}^{2} + {\left(4 - 3\right)}^{2}} = \sqrt{65} \approx 8.06$

Along the line segment connecting the two centers
$5$ units are covered by circle A, and
$1$ unit is covered by circle B.

So only $5 + 1 = 6$ units are covered by the circles.

Leaving $\sqrt{65} - 6 \approx 2.06$ units uncovered.

May 3, 2016

no overlap , ≈ 2.06

#### Explanation:

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

• If sum of radii > d , then circles overlap

• If sum of radii < d , then no overlap

To calculate the distance between the centres 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 $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 points}$

let $\left({x}_{1} , {y}_{1}\right) = \left(1 , 4\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(9 , 3\right)$

d=sqrt((9-1)^2+(3-4)^2)=sqrt65 ≈ 8.06

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

Since sum of radii < d , then no overlap

smallest distance = 8.06 - 6 = 2.06
graph{(y^2-8y+x^2-2x-8)(y^2-6y+x^2-18x+89)=0 [-35.56, 35.56, -17.78, 17.78]}