# Circle A has a radius of 1  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 30, 2016

no overlap , ≈ 4.062

#### 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

Before calculating d, we have 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 the translation $\left(\begin{matrix}- 3 \\ 4\end{matrix}\right)$

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

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

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

The 2 points here are (7 ,2) and (3 ,9)

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 = 1 + 3 = 4

Since sum of radii < d , then no overlap

min distance between points = d - sum of radii

$= 8.062 - 4 = 4.062$
graph{(y^2-4y+x^2-14x+52)(y^2-18y+x^2-6x+81)=0 [-28.48, 28.47, -14.25, 14.23]}