# Circle A has a radius of 3  and a center of (3 ,9 ). Circle B has a radius of 2  and a center of (1 ,4 ). If circle B is translated by <3 ,-1 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Sep 17, 2016

No overlap between Circle A and Circle B. Minimum distance = 1.083. See explanation and the diagram below.

#### Explanation:

Compare the distance (d) between the centres of the circles to the sum of the radii.

1) If the sum of the radii $>$d, the circles overlap.
2) If the sum of the radii $<$d, the no overlap.

The first step here is to calculate the new centre of B under the traslation, which does not change the shape of the circle only its' position.

Under a translation $< 3 , - 1 >$
$B \left(1 , 4\right) \implies \left(1 + 3 , 4 - 1\right) \implies \left(4 , 3\right)$ new centre of B

To calculate d, use the distance formula :
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

where $\left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right)$ are 2 coordinate points

here the two points are $\left(3 , 9\right)$ and $\left(4 , 3\right)$ the centres of the circles

let$\left({x}_{1} , {y}_{1}\right) = \left(3 , 9\right)$ and $\left({x}_{2} , {y}_{2}\right) = \left(4 , 3\right)$

$d = \sqrt{{\left(4 - 3\right)}^{2} + {\left(3 - 9\right)}^{2}} = \sqrt{{1}^{2} + {\left(- 6\right)}^{2}} = \sqrt{37} = 6.083$

Sum of radii = radius of A + radius of B $= 3 + 2 = 5$

Since sum of radius $<$d, then no overlap of the circles

Min.D minimum distance (the red line in the daigram) :

$d -$sum of radii $= 6.083 - 5 = 1.083$