# Circle A has a radius of 3  and a center at (1 ,3 ). Circle B has a radius of 5  and a center at (1 ,7 ). 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?

May 28, 2016

no overlap , ≈0.246

#### 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

Our first step is to find the new centre of B under the given translation.The shape of a figure does not change under a translation only it's position.

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

centre of B (1 ,7) → (1+2 ,7+4) → (3 ,11)

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 $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 points}$

Here the 2 points are (1 ,3) and (3 ,11)

d=sqrt((3-1)^2+(11-3)^2)=sqrt68≈8.246

radius of A + radius of B = 3 + 5 = 8

Since sum of radii < d , then no overlap.

minimum distance between 2 points = 8.246 - 8 = 0.246
graph{(y^2-6y+x^2-2x+1)(y^2-22y+x^2-6x+105)=0 [-20, 20, -10, 10]}