Circle A has a radius of 2  and a center at (7 ,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?

Jan 19, 2017

no overlap, min. distance ≈ 6.22

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 'new' centre of B under the given translation which does not change the shape of the circle only it's position.

Under a translation $\left(\begin{matrix}- 2 \\ 6\end{matrix}\right)$

$\left(3 , 2\right) \to \left(3 - 2 , 2 + 6\right) \to \left(1 , 8\right) \leftarrow \text{ new centre of B}$

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 $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) \text{ are 2 coordinate points}$

The 2 points here are (7 ,1) and (1 ,8)

let $\left({x}_{1} , {y}_{1}\right) = \left(7 , 1\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(1 , 8\right)$

d=sqrt((1-7)^2+(8-1)^2)=sqrt(36+49)=sqrt85≈9.22

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

min. distance between points = d - sum of radii

$\Rightarrow \text{min. distance } = 9.22 - 3 = 6.22$
graph{(y^2-2y+x^2-14x+46)(y^2-16y+x^2-2x+64)=0 [-20, 20, -10, 10]}