# Circle A has a radius of 2  and a center of (8 ,2 ). Circle B has a radius of 4  and a center of (5 ,3 ). If circle B is translated by <-1 ,5 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Aug 4, 2016

no overlap, min distance ≈ 1.211 units

#### Explanation:

What we have to do 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

Before doing this we require to find the new centre of B under the translation, which does not change the shape of the circle only it's position.

Under a translation $\left(\begin{matrix}- 1 \\ 5\end{matrix}\right)$

(5 ,3) → (5-1,3+5) → (4 ,8) is the new centre of B.

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 (8 ,2) and (4 ,8) the centres of the circles.

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

d=sqrt((4-8)^2+(8-2)^2)=sqrt(16+36)=sqrt52≈7.211