Circle A has a radius of 3  and a center of (2 ,5 ). Circle B has a radius of 3  and a center of (3 ,8 ). If circle B is translated by <4 ,2 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

May 3, 2016

no overlap , ≈ 1.07

Explanation:

What we have to do here is compare the distance (d) between the centres with the sum of the radii.

• If sum of radii > d , then circles overlap

• If sum of radii < d , then no overlap

First we require to find the new position of centre B. A translation does not change the shape of a figure only it's position.

Under a translation of $\left(\begin{matrix}4 \\ 2\end{matrix}\right)$

centre of B(3 ,8) → (3+4 ,8+2) → (7 ,10)

To calculate d , use the $\textcolor{b l u e}{\text{ distance formula }}$

color(red)(|bar(ul(color(white)(a/a)color(black)(d=sqrt((x_2-x_1)^2+(y_2-y_1)^2))color(white)(a/a)|))
where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 points}$

let $\left({x}_{1} , {y}_{1}\right) = \left(2 , 5\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(7 , 10\right)$

d=sqrt((7-2)^2+(10-5)^2)=sqrt50 ≈ 7.07

radius of A + radius of B = 3 + 3 = 6

Since sum of radii < d , then no overlap

and minimum distance = 7.06 - 6 = 1.06
graph{(y^2-10y+x^2-4x+20)(y^2-20y+x^2-14x+140)=0 [-35.56, 35.56, -17.78, 17.78]}