# Circle A has a radius of 6  and a center of (2 ,5 ). Circle B has a radius of 3  and a center of (6 ,7 ). If circle B is translated by <3 ,1 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Aug 9, 2016

circles overlap.

#### Explanation:

What we require 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

Before doing this we need 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 the translation $\left(\begin{matrix}3 \\ 1\end{matrix}\right)$

(6 ,7) → (6+3 ,7+1) → (9 ,8) is the 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 $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points}$

Here the 2 points are (2 ,5) and (9 ,8) the centres of the circles.

let $\left({x}_{1} , {y}_{1}\right) = \left(2 , 5\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(9 , 8\right)$

d=sqrt((9-2)^2+(8-5)^2)=sqrt(49+9)=sqrt58≈7.616

sum of radii = radius of A + radius of B = 6 + 3 = 9

Since sum of radii > d , then circles overlap
graph{(y^2-10y+x^2-4x-7)(y^2-16y+x^2-18x+136)=0 [-40, 40, -20, 20]}