Circle A has a radius of 2  and a center of (7 ,6 ). Circle B has a radius of 3  and a center of (2 ,3 ). 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?

Jan 28, 2017

no overlap , min. distance ≈1.083 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 $\textcolor{b l u e}{\text{sum of radii}}$

• If sum of radii > d , then circles overlap

• If sum of radii < d , then no overlap of circles

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

$\left(2 , 3\right) \to \left(2 - 1 , 3 + 2\right) \to \left(1 , 5\right) \leftarrow \text{ new centre of B}$

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 ,6) and (1 ,5)

let $\left({x}_{1} , {y}_{1}\right) = \left(7 , 6\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(1 , 5\right)$

d=sqrt((1-7)^2+(5-6)^2)=sqrt(36+1)=sqrt37≈6.083

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

Since sum of radii < d, then no overlap of circles

min. distance between points = d - sum of radii

$= 6.083 - 5 = 1.083$
graph{(y^2-12y+x^2-14x+81)(y^2-10y+x^2-2x+17)=0 [-20, 20, -10, 10]}