Circle A has a radius of 2  and a center of (7 ,2 ). Circle B has a radius of 3  and a center of (5 ,7 ). If circle B is translated by <-1 ,2 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Oct 30, 2016

no overlap , ≈ 2.616 units.

Explanation:

What we have to do here is $\textcolor{b l u e}{\text{compare}}$ the distance (d) between the centres of the circles to the $\textcolor{b l u e}{\text{sum of the radii}}$

• If sum of radii > d , then circles overlap

• If sum of radii < d , then no overlap

Before calculating d, we require to find the coordinates of the ' new' centre of circle B under the given translation which does not change the shape of the circle, only it's position.

Under a translation $\left(\begin{matrix}- 1 \\ 2\end{matrix}\right)$

$B \left(5 , 7\right) \to \left(5 - 1 , 7 + 2\right) \to \left(4 , 9\right) \leftarrow \text{ new centre of B}$

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

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points}$

The 2 points here are (7 ,2) and (4 ,9)

let $\left({x}_{1} , {y}_{1}\right) = \left(7 , 2\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(4 , 9\right)$

d=sqrt((4-7)^2+(9-2)^2)=sqrt(9+49)=sqrt58≈7.616

Sum of radii = radius of A + radius of B = 2 + 3 = 5

Since sum of radii < d, then no overlap

and min. distance between points = d - sum of radii

$= 7.616 - 5 = 2.616$
graph{(y^2-4y+x^2-14x+49)(y^2-18y+x^2-8x+88)=0 [-25.31, 25.32, -12.66, 12.65]}