Circle A has a radius of 3  and a center of (2 ,1 ). Circle B has a radius of 2  and a center of (7 ,3 ). If circle B is translated by <4 ,2 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Oct 2, 2016

no overlap , ≈ 4.849

Explanation:

What we have to do here is $\textcolor{b l u e}{\text{compare}}$ the distance ( d) between the centres of the circles to the $\textcolor{b l u e}{\text{sum of the radii}}$

• If sum of radii > d , then circles overlap

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

Before calculating d, we require to find the 'new' centre of B under the given translation which does not change the shape of the circle only it's position.

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

7,3)to(7+4,3+2)to(11,5)larr" new centre of 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 $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points}$

The 2 points here are (2 ,1) and (11 ,5)

let $\left({x}_{1} , {y}_{1}\right) = \left(2 , 1\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(11 , 5\right)$

d=sqrt((11-2)^2+(5-1)^2)=sqrt(81+16)=sqrt97≈9.849

Sum of radii = radius of A + radius of B = 3 + 2 = 5

Since sum of radii < d , then no overlap

min. distance between circles = d - sum of radii

$= 9.849 - 5 = 4.849$
graph{(y^2-2y+x^2-4x-4)(y^2-10y+x^2-22x+142)=0 [-20, 20, -10, 10]}