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

Jul 8, 2016

circle B overlaps circle A

Explanation:

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

• If sum of radii > d , then circles overlap

• If sum of radii < d , then no overlap

The first step,however, is to find the coordinates of the centre of circle B under the translation.

Under a translation of $\left(\begin{matrix}- 2 \\ 3\end{matrix}\right)$

B(5 ,1) → (5-2 ,1+3) → B(3 ,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 $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points}$

Here the 2 points are A(1 ,2) and B(3 ,4)

let $\left({x}_{1} , {y}_{1}\right) = \left(1 , 2\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(3 , 4\right)$

d=sqrt((3-1)^2+(4-2)^2)=sqrt(4+4)=sqrt8≈2.828

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

Since sum of radii > d , then circles overlap.
graph{(y^2-4y+x^2-2x+4)(y^2-8y+x^2-6x+21)=0 [-13.86, 13.86, -6.92, 6.94]}