Question

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

Answer 1

The circle B does not overlap circle A.
The minimum distance between points on both circles is 3 units

Explanation:

Circle A:
radius `r_1=2`
center `A-=(5,1)`
Circle B:
radius `r_2=1`
center `B-=(3,2)`
translation of B `B'-=<-2,6>`
new center `(3-2,2+6)=(5,8)`
Distance between center of A and new center of B' is
`sqrt((5-5)^2+(8-1)^2)=sqrt(0^2+7^2)=sqrt7^2=7`
`AB'=7`
radius of A `r_1=2`
radius of B `r_2=1`

Sum of the radii `r_1+r_2=2+1=3`

Sum of the radii `<` Distance between centers A and B.
Hence, they do d
not overlap.
The difference between
the distance between the centers'an
the sum of the radii
represents the shortest distance between the two circles

separation `=7-3=4`

The two circles are separated by 3 units

Answer 2

`"no overlap "~~5.062`

Explanation:

`"what we have to do here is "color(blue)"compare ""the distance (d)"`
`"between the centres to the "color(blue)"sum of 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 centre of"`
`"B under the given translation"`

`"under the translation "<-2,6>`

`(3,2)to(3-2,2+6)to(1,8)larrcolor(red)"new centre of B"`

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

`•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`"let "(x_1,y_1)=(5,1)" and "(x_2,y_2)=(1,8)`

`d=sqrt((1-5)^2+(8-1)^2)=sqrt(16+49)=sqrt65~~8.062`

`"sum of radii "=2+1=3`

`"since sum of radii"< d" then no overlap"`

`"min. distance "=d-" sum of radii"`

`color(white)(xxxxxxxxxx)=8.062-3=5.062`
graph{((x-5)^2+(y-1)^2-4)((x-1)^2+(y-8)^2-1)=0 [-20, 20, -10, 10]}