Question

Circle A has a center at `(1 ,-2 )` and a radius of `3`. Circle B has a center at `(-4 ,-8 )` and a radius of `2`. Do the circles overlap? If not, what is the smallest distance between them?

Answer 1

Smallest distance between the two circles is `sqrt(61) - 5 ~~2.81`.

Explanation:

Let's compute the distance between the two centers of the circle.

This can be done with Pythagorean theorem:

`d^2 = d_x^2 + d_y^2`

where `d` is the distance between the two points, `d_x` is the distance between the `x` values and `d_y` is the distance between the `y` values of the points.

In your case, you have

`d^2 = (1 -(-4))^2 + (-2 - (-8))^2 = 25 + 36 = 61`

`=> d = sqrt(61) ~~ 7.81`

However, `7.81` is the distance between the centers and not the distance between the outer points of the circle.
To compute the distance between the circles, we also need to take the radius `r_1 = 3` and `r_2 = 3` into consideration:

`"smallest distance" = d - r_1 - r_2 = d - 3 - 2 = sqrt(61) - 5 ~~ 2.81`

Thus, the smallest distance between these two circles is `~~2.81`.

I will insert a graph of the two circles with the line that passes through both centers.
The smallest distance is the length of this line between the two circles.

graph{((x-1)^2 + (y+2)^2 - 9)((x+4)^2 + (y+8)^2 - 4)(y - (6/5 x - 16/5)) = 0 [-13.92, 11.4, -10.73, 1.93]}