Question

Circle A has a center at `(5 ,4 )` and an area of `4 pi`. Circle B has a center at `(2 ,8 )` and an area of `9 pi`. Do the circles overlap? If not, what is the shortest distance between them?

Answer 1

The circles touch, but they do not overlap. The shortest distance between them is `0`.

Explanation:

The distance between the points `(5,4)` and `(2,8)` is
`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
`d=sqrt((2-5)^2+(8-4)^2)`
`d=sqrt((-3)^2+4^2)`
`d=sqrt(9+16)`
`d=sqrt(25)=5`

If the circles overlap, the sum of their radii will be greater than `d=5.` If they don't, the shortest distance between them will be what we get when we subtract both radii from `5`.

Let `r_1` and `r_2` be the radius of circles 1 and 2, respectively. Then:

`A_1=pir_1^2`
`=>r_1=sqrt(A_1/pi)=sqrt((4cancelpi)/cancelpi)=sqrt(4)=2`

Similarly,
`r_2=sqrt(A_2/pi)=sqrt((9cancelpi)/cancelpi)=sqrt(9)=3`

We get
`"     "r_1+r_2`
`=2+3`
`=5"           "=d`

Since the sum of the radii matches the distance between the two circles' centers, the circles touch, but they do not overlap. The shortest distance between the two circles is `0`.