Circle A has a radius of 1  and a center at (7 ,4 ). Circle B has a radius of 3  and a center at (6 ,5 ). If circle B is translated by <-3 ,4 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Dec 21, 2017

$\text{no overlap } \approx 2.403$

Explanation:

What we have to do here is $\textcolor{b l u e}{\text{compare}}$ the distance between the centres of the circles with 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"

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

$\text{under a translation } < - 3 , 4 >$

$\left(6 , 5\right) \to \left(6 - 3 , 5 + 4\right) \to \left(3 , 9\right) \leftarrow \textcolor{red}{\text{new centre of B}}$

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

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

$\text{let "(x_1,y_1)=(7,4)" and } \left({x}_{2} , {y}_{2}\right) = \left(3 , 9\right)$

$d = \sqrt{{\left(3 - 7\right)}^{2} + {\left(9 - 4\right)}^{2}} = \sqrt{41} \approx 6.403$

$\text{sum of radii } = 1 + 3 = 4$

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

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

$\textcolor{w h i t e}{\times \times \times \times \times \times \times} = 6.403 - 4 = 2.403$
graph{((x-7)^2+(y-4)^2-1)((x-3)^2+(y-9)^2-9)=0 [-20, 20, -10, 10]}