# Circle A has a radius of 3  and a center of (3 ,2 ). Circle B has a radius of 5  and a center of (4 ,7 ). If circle B is translated by <2 ,-1 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Sep 27, 2016

circles overlap.

#### 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 $\left(\begin{matrix}2 \\ - 1\end{matrix}\right)$

$\left(4 , 7\right) \to \left(4 + 2 , 7 - 1\right) \to \left(6 , 6\right) \leftarrow \text{ new centre of B}$

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

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points}$

The 2 points here are (3 ,2) and (6 ,6)

let $\left({x}_{1} , {y}_{1}\right) = \left(3 , 2\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(6 , 6\right)$

$d = \sqrt{{\left(6 - 3\right)}^{2} + {\left(6 - 2\right)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$