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

May 21, 2017

The new centre of circle B will be $\left(3 , 5\right)$, which is $4.472$ units from the centre of circle A. Since the sum of the radii of the two circles is $5$ units, the circles overlap.

#### Explanation:

Translating circle B by $< 1 , 3 >$ simply requires adding 1 to the x-value and 3 to the y-value of the coordinates of its centre, so the new centre of circle B is $\left(3 , 5\right)$.

The radii of the circles are 2 and 3 respectively, so if their centres are now less than 5 units apart they will overlap, but if they are more than 5 units apart they will not.

To find the distance between the centres, $r$, we use an application of Pythagoras Theorem:

$r = \sqrt{{\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({x}_{2} - {x}_{1}\right)}^{2}} = \sqrt{{\left(5 - 3\right)}^{2} + {\left(3 - 7\right)}^{2}}$

$= \sqrt{{\left(2\right)}^{2} + {\left(- 4\right)}^{2}} = \sqrt{4 + 16} = \sqrt{20} = 4.472$ units

Because this is more than $5$ units, the circles overlap. And hence, the question of the minimum distance between the circles does not arise.

graph{((x-3)^2+(y-5)^2-9)((x-7)^2+(y-3)^2-4)((x-2)^2+(y-2)^2-0.02)((x-3)^2+(y-5)^2-0.02)=0 [-6.04, 13.96, -1.44, 8.56]}