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

Jul 24, 2016

circles overlap.

#### Explanation:

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

• If sum of radii > d , then circles overlap

• If sum of radii < d , then no overlap

The first step is to calculate the coordinates of the 'new' centre of circle B under the given translation. Under a translation the circle remains a circle but it's position changes.

Under a translation $\left(\begin{matrix}2 \\ 4\end{matrix}\right)$

(1 ,2) → (1+2 ,2+4) → new centre of B is (3 ,6)

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 (5 ,3) and (3 ,6) the centres of the circles.

let $\left({x}_{1} , {y}_{1}\right) = \left(5 , 3\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(3 , 6\right)$

d=sqrt((3-5)^2+(6-3)^2)=sqrt(4+9)=sqrt13≈3.606

sum of radii = radius of A + radius of B = 4 + 2 = 6

Since sum of radii > d , then circles overlap
graph{(y^2-6y+x^2-10x+18)(y^2-12y+x^2-6x+41)=0 [-20, 20, -10, 10]}