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

Aug 6, 2016

no overlap , min. distance ≈ 0.082 units

Explanation:

What we have to do here is to compare the distance ( d) between the centres of the circles with 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 require to find the new coordinates of the centre of B under the translation, which does not change the shape of the circle, only it's position.

Under a translation $\left(\begin{matrix}2 \\ 7\end{matrix}\right)$

(6 ,1) → (6+2 ,1+7) → (8 ,8) is the new centre of circle 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 ,7) and (8 ,8) the centres of the circles.

let $\left({x}_{1} , {y}_{1}\right) = \left(2 , 7\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(8 , 8\right)$

d=sqrt((8-2)^2+(8-7)^2)=sqrt(3 6+1)=sqrt37≈6.082

sum of radii = radius of A + radius of B = 5 + 1 = 6

Since sum of radii < d , then no overlap

min. distance between points = d - sum of radii

= 6.082 - 6 = 0.082 units (3 decimal paces)
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