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

Jul 3, 2016

no overlap , minimum distance ≈ 0.61

#### 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 the sum of radii > d , then circles overlap

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

Before calculating d , we require to find the new centre of B under the translation.

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

B(8 ,1) → B(8-4 ,1+3) → B(4 ,4)

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 A(1 ,2) and B(4 ,4)

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

d=sqrt((4-1)^2+(4-2)^2)=sqrt(9+4)=sqrt13≈3.61

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

Since sum of radii < d , then no overlap

Minimum distance between 2 points on circles is.

d - sum of radii = 3.61 - 3 = 0.61
graph{(y^2-4y+x^2-2x+4)(y^2-8y+x^2-8x+28)=0 [-12.65, 12.66, -6.33, 6.32]}