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

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Jim G. Share
Jan 26, 2018

#### Answer:

$\text{no overlap}$

#### Explanation:

What we have to do here is $\textcolor{b l u e}{\text{compare }}$ thedistance (d) between the centres with 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"

$\text{before we can calculate d we require the 'new' centre of}$
$\text{circle B}$

$\text{under a translation of } < - 3 , 4 >$

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

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

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

$\text{let "(x_1,y_1)=(6,1)" and } \left({x}_{2} , {y}_{2}\right) = \left(1 , 9\right)$

$d = \sqrt{{\left(1 - 6\right)}^{2} + {\left(9 - 1\right)}^{2}} = \sqrt{25 + 64} \approx 9.43$

$\text{sum of radii } = 5 + 1 = 6$

$\text{since sum of radii"< d" then no overlap}$

$\text{minimum distance "=d-" sum of radii}$

$\textcolor{w h i t e}{\text{minimum distance }} = 9.43 - 6 = 3.43$
graph{((x-6)^2+(y-1)^2-25)((x-1)^2+(y-9)^2-1)=0 [-20, 20, -10, 10]}

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