# Circle A has a radius of 1  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 <-2 ,5 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Jan 10, 2017

no overlap, min. distance ≈ 2.81

#### Explanation:

What we have to do here is $\textcolor{b l u e}{\text{compare}}$ the distance (d ) between the centres of the circles to the $\textcolor{b l u e}{\text{sum of the radii}} .$

• If sum of radii > d , then circles overlap

• If sum of radii < d , then no overlap

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

Under the translation $\left(\begin{matrix}- 2 \\ 5\end{matrix}\right)$

$\left(5 , 3\right) \to \left(5 - 2 , 3 + 5\right) \to \left(3 , 8\right) \leftarrow \text{ new centre of B}$

To calculate d, use the $\textcolor{b l u e}{\text{distance formula}}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) \text{ are 2 coordinate points}$

The 2 points here are (8 ,2) and (3 ,8)

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

d=sqrt((3-8)^2+(8-2)^2)=sqrt(25+36)=sqrt61≈7.81

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

Since sum of radii < d , then there is no overlap

min. distance between points = d - sum of radii

$= 7.81 - 5 = 2.81$
graph{(y^2-4y+x^2-16x+67)(y^2-16y+x^2-6x+57)=0 [-32.05, 32.03, -16, 16.04]}