Question

Circle A has a radius of `6` and a center of `(4 ,3 )`. Circle B has a radius of `3` and a center of `(1 ,8 )`. If circle B is translated by `<-2 ,4 >`, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Answer 1

no overlap, min distance ≈1.296

Explanation:

What we have to do here is `color(blue)"compare"` the distance (d ) between the centres of the circles to the `color(blue)"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 under the given translation which does not change the shape of the circle only it's position.

`"Under a translation "((-2),(4))`

`(1,8)to(1-2,8+4)to(-1,12)larrcolor(blue)" new centre of B"`

To calculate d, use the `color(blue)"sistance formula"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(d=sqrt((x_2-x_1)^2+(y_2-y_1)^2))color(white)(2/2)|)))`
where `(x_1,y_1),(x_2,y_2)" are 2 coordinate points"`

The 2 points here are (4 ,3) and (-1 ,12)

let `(x_1,y_1)=(4,3)" and " (x_2,y_2)=(-1,12)`

`d=sqrt((-1-4)^2+(12-3)^2)=sqrt(25+81)≈10.296`

Sum of radii = radius of A + radius of B = 6 + 3 = 9

Since sum of radii < d , then circles do not overlap

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

`rArr"min. distance "=10.296-9=1.296`
graph{((x-4)^2+(y-3)^2-36)((x+1)^2+(y-12)^2-9)=0 [-56.2, 56.2, -28.1, 28.1]}