# Circle A has a radius of 4  and a center of (6 ,2 ). Circle B has a radius of 3  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?

Jun 1, 2018

$\text{no overlap } \approx 0.62$

#### Explanation:

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

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

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

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

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

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

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

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

$\text{let "(x_1,y_1)=(6,2)" and } \left({x}_{2} , {y}_{2}\right) = \left(3 , 9\right)$

$d = \sqrt{{\left(3 - 6\right)}^{2} + {\left(9 - 2\right)}^{2}} = \sqrt{9 + 49} = \sqrt{58} \approx 7.62$

$\text{sum of radii } = 4 + 3 = 7$

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

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

$\textcolor{w h i t e}{\times \times \times \times \times \times x} = 7.62 - 7 = 0.62$
graph{((x-6)^2+(y-2)^2-16)((x-3)^2+(y-9)^2-9)=0 [-20, 20, -10, 10]}