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

May 19, 2018

$\text{no overlap ",~~1.39" to 2 dec. places}$

Explanation:

$\text{What we have to do here is compare the distance (d) }$
$\text{between the centres to the sum of their radii}$

• " if sum of radii">d" then circles overlap"

• " if sum of radii">d" then no overlap"

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

$\text{under the translation } < 2 , - 3 >$

$\left(4 , 7\right) \to \left(4 + 2 , 7 - 3\right) \to \left(6 , 4\right) \leftarrow \textcolor{red}{\text{new centre of B}}$

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

•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

$\text{let "(x_1,y_1)=(6,4)" and } \left({x}_{2} , {y}_{2}\right) = \left(1 , 2\right)$

$d = \sqrt{{\left(1 - 6\right)}^{2} + {\left(2 - 4\right)}^{2}} = \sqrt{25 + 4} = \sqrt{29} \approx 5.39$

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

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

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

$\textcolor{w h i t e}{\text{minimum distance }} = 5.39 - 4 = 1.39$
graph{((x-1)^2+(y-2)^2-9)((x-6)^2+(y-4)^2-1)=0 [-10, 10, -5, 5]}