# Circle A has a radius of 2  and a center of (4 ,6 ). Circle B has a radius of 3  and a center of (9 ,8 ). 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?

Mar 27, 2017

no overlap, min. distance ≈ 1.708 units.

#### 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 there is no overlap

Before calculating d we require to find the 'new' centre of B under the given translation which does no change the shape of the circle only it's position.

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

$\left(9 , 8\right) \to \left(9 - 2 , 8 + 4\right) \to \left(7 , 12\right) \rightarrow \textcolor{red}{\text{ new centre of B}}$

To calculate d, use the $\textcolor{b l u e}{\text{distance formula}}$

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

The 2 points here are (4 ,6) and (7 ,12)

let $\left({x}_{1} , {y}_{1}\right) = \left(4 , 6\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(7 , 12\right)$

$d = \sqrt{{\left(7 - 4\right)}^{2} + {\left(12 - 6\right)}^{2}} = \sqrt{9 + 36} = \sqrt{45} \approx 6.708$

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

Since sum of radii < d, then no overlap

min. distance = d - sum of radii

$\Rightarrow \text{ min. distance } = 6.708 - 5 = 1.708$
graph{(y^2-12y+x^2-8x+48)(y^2-24y+x^2-14x+184)=0 [-40, 39.98, -20, 19.98]}