Circle A has a radius of 5  and a center of (6 ,2 ). Circle B has a radius of 1  and a center of (4 ,5 ). If circle B is translated by <-3 ,4 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Oct 3, 2016

no overlap, ≈ 2.602

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

Under a translation $\left(\begin{matrix}- 3 \\ 4\end{matrix}\right)$

$\left(4 , 5\right) \to \left(4 - 3 , 5 + 4\right) \to \left(1 , 9\right) \leftarrow \text{ centre of circle B}$

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

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

The 2 points here are (6 ,2) and (1 ,9)

let $\left({x}_{1} , {y}_{1}\right) = \left(6 , 2\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(1 , 9\right)$

d=sqrt((1-6)^2+(9-2)^2)=sqrt(25+49)=sqrt74≈8.602

Sum of radii = radius of A + radius of B = 5 + 1 = 6

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

min. distance between them = d - sum of radii

$= 8.602 - 6 = 2.602$
graph{(y^2-4y+x^2-12x+15)(y^2-18y+x^2-2x+81)=0 [-20, 20, -10, 10]}