Question

Circle A has a radius of `2` and a center of `(6 ,5 )`. Circle B has a radius of `3` and a center of `(2 ,4 )`. If circle B is translated by `<1 ,1 >`, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Answer 1

`"circles overlap"`

Explanation:

`"what we have to do here is compare the distance (d)"`
`"between the centres to the sum of the radii"`

`• " if sum of radii">d" then circles overlap"`

`• " if sum of radii"< d" then no overlap"`

`"before calculating d we require to find the new centre"`
`"of B after the given translation"`

`"under the translation "< 1,1>`

`(2,4)to(2+1,4+1)to(3,5)larrcolor(red)"new centre of B"`

`"to calculate d use the "color(blue)"distance formula"`

`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`"let "(x_1,y_1)=(6,5)" and "(x_2,y_2)=(3,5)`

`d=sqrt((3-6)^2+(5-5)^2)=sqrt9=3`

`"sum of radii "=2+3=5`

`"since sum of radii">d" then circles overlap"`
graph{((x-6)^2+(y-5)^2-4)((x-3)^2+(y-5)^2-9)=0 [-20, 20, -10, 10]}

Answer 2

The distance between the centers is `3`, which satisfies the triangle inequality with the two radii of `2` and `3`, so we have overlapping circles.

Explanation:

I thought I did this one already.

A is `(6,5)` radius `2`

B's new center is `(2,4)+<1,1> =(3,5),` radius still `3`

Distance between centers,

`d = sqrt{ (6-3)^2 + (5-5)^2}=3`

Since the distance between the centers is less than the sum of the two radii, we have overlapping circles.