Circle A has a radius of 3  and a center at (1 ,2 ). Circle B has a radius of 5  and a center at (3 ,4 ). 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?

Jan 18, 2018

$\text{circles overlap}$

Explanation:

What we have to do here is $\textcolor{b l u e}{\text{compare }}$the distance (d) between the centres to the $\textcolor{b l u e}{\text{sum of radii}}$

• " if "a>d" then circles overlap"

• " if "a < d" then no overlap"

$\text{before we can calculate d we require to find the new }$
$\text{centre of B under the given translation}$

$\text{under the translation } < 2 , 1 >$

$\left(3 , 4\right) \to \left(3 + 2 , 4 + 1\right) \to \left(5 , 5\right) \leftarrow \textcolor{red}{\text{new centre of B}}$

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

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

$\text{let "(x_1,y_1)=(1,2)" and } \left({x}_{2} , {y}_{2}\right) = \left(5 , 5\right)$

$d = \sqrt{{\left(5 - 1\right)}^{2} + {\left(5 - 2\right)}^{2}} = \sqrt{16 + 9} = 5$

$\text{sum of radii } = 3 + 5 = 8$

$\text{since sum of radii ">d" then circles overlap}$
graph{((x-1)^2+(y-2)^2-9)((x-5)^2+(y-5)^2-25)=0 [-20, 20, -10, 10]}