Question

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?

Answer 1

The new centre of circle B will be `(3,5)`, 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 `(3,5)`.

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((y_2-y_1)^2+(x_2-x_1)^2) = sqrt((5-3)^2+(3-7)^2)`

`= sqrt((2)^2+(-4)^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]}