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

Aug 31, 2016

no overlap, ≈ 0.657

#### Explanation:

What we have to do here is compare the distance ( d) between the centres of the circles to the sum of the radii.

•If sum of radii > d , then circles overlap

• If sum of radii < d , then no overlap

The first step, however , is 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 $\left(\begin{matrix}- 3 \\ 2\end{matrix}\right)$

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

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 $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points}$

The 2 points here are (6 ,1) and (2 ,5) the centres of the circles.

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

d=sqrt((2-6)^2+(5-1)^2)=sqrt(16+16)=sqrt32≈5.657

sum of radii = radius of A + radius of B = 4 + 1 = 5

Since sum of radii < d , then no overlap and min. distance between circles is.

min. distance = d - sum of radii = 5.657 - 5 = 0.657 units.
graph{(y^2-2y+x^2-12x+21)(y^2-10y+x^2-4x+28)=0 [-20, 20, -10, 10]}