Circle A has a radius of 4  and a center of (6 ,2 ). Circle B has a radius of 2  and a center of (5 ,7 ). If circle B is translated by <-2 ,2 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

May 5, 2017

$\text{no overlap } , \approx 1.616$

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"

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

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

$\left(5 , 7\right) \to \left(5 - 2 , 7 + 2\right) \to \left(3 , 9\right) \leftarrow \textcolor{red}{\text{ new centre of B}}$

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

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

$\text{where " (x_1,y_1),(x_2,y_2)" are 2 coordinate points}$

$\text{the 2 points here are " (6,2)" and } \left(3 , 9\right)$

$d = \sqrt{{\left(3 - 6\right)}^{2} + {\left(9 - 2\right)}^{2}} = \sqrt{9 + 49} = \sqrt{58} \approx 7.616$

$\text{sum of radii } = 4 + 2 = 6$

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

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

$\Rightarrow \text{min distance } = 7.616 - 6 = 1.616$
graph{(y^2-4y+x^2-12x+24)(y^2-18y+x^2-6x+86)=0 [-28.87, 28.86, -14.43, 14.44]}