Circle A has a radius of 2  and a center of (6 ,6 ). Circle B has a radius of 3  and a center of (2 ,4 ). If circle B is translated by <1 ,5 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Oct 7, 2017

$\text{circles overlap}$

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 $\textcolor{b l u e}{\text{sum of radii}}$

• " if sum of radii">d" then circles overlap"

• " if sum of radii"< d" then no overlap"

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.

$\text{under the translation } \left(\begin{matrix}1 \\ 5\end{matrix}\right)$

$\left(2 , 4\right) \to \left(2 + 1 , 4 + 5\right) \to \left(3 , 9\right) \leftarrow \textcolor{red}{\text{ new centre of B}}$

$\text{to calculate d use the "color(blue)"distance formula}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{let "(x_1,y_1)=(6,6)" and } \left({x}_{2} , {y}_{2}\right) = \left(3 , 9\right)$

$d = \sqrt{{\left(3 - 6\right)}^{2} + {\left(9 - 6\right)}^{2}} = \sqrt{18} \approx 4.243$

$\text{sum of radii } = 2 + 3 = 5$

$\text{since sum of radii">d" then circles overlap}$
graph{((x-6)^2+(y-6)^2-4)((x-3)^2+(y-9)^2-9)=0 [-20, 20, -10, 10]}