# Circle A has a radius of 3  and a center of (2 ,1 ). Circle B has a radius of 2  and a center of (7 ,8 ). If circle B is translated by <4 ,2 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Aug 23, 2017

$\text{no overlap} \approx 7.727$

#### 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 $\textcolor{b l u e}{\text{sum of radii}}$

• " 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.

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

$\left(7 , 8\right) \to \left(7 + 4 , 8 + 2\right) \to \left(11 , 10\right) \leftarrow \textcolor{red}{\text{ new centre of B}}$

$\text{to calculate d use the "color(blue)"distance formula}$

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

$\left({x}_{1} , {y}_{1}\right) = \left(2 , 1\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(11 , 10\right)$

$d = \sqrt{{\left(11 - 2\right)}^{2} + {\left(10 - 1\right)}^{2}} = \sqrt{162} \approx 12.727$

$\text{sum of radii } = 3 + 2 = 5$

$\text{since sum of radii "< d " then no overlap}$

$\text{minimum distance "=d-" sum of radii}$

$\textcolor{w h i t e}{\min i \mu m \mathrm{di} s \tan c e} = 12.727 - 5 = 7.727$
graph{((x-2)^2+(y-1)^2-9)((x-11)^2+(y-10)^2-4)=0 [-20, 20, -10, 10]}