# Circle A has a radius of 1  and a center of (5 ,4 ). Circle B has a radius of 2  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?

Apr 21, 2016

no overlap , ≈ 3.4

#### Explanation:

Here, we have to calculate the distance (d) between the centres and compare this with the sum of the radii.

• If sum of radii > d , then circles overlap.

• If sum of radii < d , then no overlap.

Now a translation does not change the shape of a figure , only it's position.

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

centre of B(4 , 5) → (4-3 , 5+4) → (1 , 9)

To calculate the distance between the centres , 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 $\left({x}_{1} , {y}_{1}\right) = \left(5 , 4\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(1 , 9\right)$

 d =sqrt((1-5)^2+(9-4)^2)=sqrt(16+25)=sqrt41 ≈ 6.4