# Circle A has a radius of 6  and a center of (4 ,3 ). Circle B has a radius of 3  and a center of (1 ,8 ). If circle B is translated by <-2 ,4 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Mar 9, 2017

no overlap, min distance ≈1.296

#### 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.

$\text{Under a translation } \left(\begin{matrix}- 2 \\ 4\end{matrix}\right)$

$\left(1 , 8\right) \to \left(1 - 2 , 8 + 4\right) \to \left(- 1 , 12\right) \leftarrow \textcolor{b l u e}{\text{ new centre of B}}$

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

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

The 2 points here are (4 ,3) and (-1 ,12)

let $\left({x}_{1} , {y}_{1}\right) = \left(4 , 3\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(- 1 , 12\right)$

d=sqrt((-1-4)^2+(12-3)^2)=sqrt(25+81)≈10.296

Sum of radii = radius of A + radius of B = 6 + 3 = 9

Since sum of radii < d , then circles do not overlap

$\text{min. distance between points "=d-" sum of radii}$

$\Rightarrow \text{min. distance } = 10.296 - 9 = 1.296$
graph{((x-4)^2+(y-3)^2-36)((x+1)^2+(y-12)^2-9)=0 [-56.2, 56.2, -28.1, 28.1]}