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

May 10, 2016

no overlap , ≈ 6.849

#### Explanation:

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

• If sum of radii > d , then circles overlap

• If sum of radii < d , then no overlap

Firstly ,we require to find the new centre of B under the translation.
A translation does not change the shape of a figure only it's position.

Under a translation of $\left(\begin{matrix}- 2 \\ 6\end{matrix}\right)$

centre B(3 ,4) → (3-2 ,4+6) → (1 ,10)

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

let (x_1,y_1)=(5,1)" and (x_2,y_2)=(1,10)

d=sqrt((1-5)^2+(10-1)^2)=sqrt(16+81)=sqrt97 ≈ 9.849

radius of A + radius of B = 2 + 1 = 3

Since sum of radii < d , then no overlap

min. distance between points = 9.849 - 3 = 6.849
graph{(y^2-2y+x^2-10x+22)(y^2-20y+x^2-2x+100)=0 [-40, 40, -20, 20]}