# Circle A has a radius of 1  and a center of (8 ,2 ). Circle B has a radius of 4  and a center of (5 ,3 ). If circle B is translated by <-1 ,5 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Aug 28, 2016

no overlap,min distance ≈ 2.211

#### Explanation:

What we have 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 calculating d, we require to find the centre of B under the given translation, which does not change the shape of the circle only it's position.

Under the translation $\left(\begin{matrix}- 1 \\ 5\end{matrix}\right)$

(5 ,3) → (5-1 ,3+5) → (4 ,8) is 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 (8 ,2) and (4 ,8) the centres of the circles.

let $\left({x}_{1} , {y}_{1}\right) = \left(8 , 2\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(4 , 8\right)$

d=sqrt((4-8)^2+(8-2)^2)=sqrt(16+36)=sqrt52≈7.211

sum of radii = radius of A + radius of B = 1 + 4 = 5

Since sum of radii < d , then no overlap of circles

min. distance between circles = d - sum of radii

$= 7.211 - 5 = 2.211$
graph{(y^2-4y+x^2-16x+67)(y^2-16y+x^2-8x+64)=0 [-24.92, 25.03, -12.48, 12.5]}