Circle A has a radius of 4  and a center at (7 ,2 ). Circle B has a radius of 3  and a center at (6 ,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?

Sep 13, 2016

no overlap , ≈ 1.062

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

However, the first step here is to calculate 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 $\left(\begin{matrix}- 3 \\ 4\end{matrix}\right)$

$B \left(6 , 5\right) \to \left(6 - 3 , 5 + 4\right) \to \left(3 , 9\right) \text{ new centre of B}$

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

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

here the 2 points are (7 ,2) and (3 ,9) the centres of the circles.

let $\left({x}_{1} , {y}_{1}\right) = \left(7 , 2\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(3 , 9\right)$

d=sqrt((3-7)^2+(9-2)^2)=sqrt(16+49)=sqrt65≈8.062

sum of radii = radius of A + radius of B = 4 + 3 = 7

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

min. distance = d - sum of radii = 8.062 - 7 = 1.062
graph{(y^2-4y+x^2-14x+37)(y^2-18y+x^2-6x+81)=0 [-40, 40, -20, 20]}