Circle A has a radius of 2  and a center of (5 ,7 ). Circle B has a radius of 4  and a center of (7 ,2 ). If circle B is translated by <2 ,-5 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Apr 16, 2016

no overlap, ≈ 4.77

Explanation:

To determine this , we require to compare the distance (d) between the centres of the circles with the sum of their radii.

• If sum of radii > d , then circles will overlap.

• If sum of radii < d , then no overlap will occur.

A translation does not change the shape of a figure , only it's position.

Under a translation of $\left(\begin{matrix}2 \\ - 5\end{matrix}\right)$

centre of B (7 , 2) → (7+2 ,2-5 ) → (9 , -3)

To calculate the distance between the centres , 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 $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points }$

let $\left({x}_{1} , {y}_{1}\right) = \left(5 , 7\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(9 , - 3\right)$

rArr d = sqrt((9-5)^2 + (-3-7)^2) =sqrt(16+100) ≈ 10.77

now, radius of A + radius of B = 2 + 4 = 6

Since sum of radii < d , no overlap

and distance between circles = 10.77 - 6 = 4.77