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

Nov 18, 2016

no overlap , ≈ 0.162

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

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 $\left(\begin{matrix}1 \\ - 4\end{matrix}\right)$

$\left(4 , 7\right) \to \left(4 + 1 , 7 - 4\right) \to \left(5 , 3\right) \leftarrow \text{ new centre of B}$

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 $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points}$

The 2 points here are (2 ,4) and (5 ,3)

let $\left({x}_{1} , {y}_{1}\right) = \left(2 , 4\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(5 , 3\right)$

d=sqrt((5-2)^2+(3-4)^2)=sqrt(9+1)=sqrt10≈3.162

Sum of radii = 1 + 2 = 3

Since sum of radii < d, then there is no overlap

min. distance between points = d - sum of radii

$= 3.162 - 3 = 0.162 \text{ to 3 decimal places}$
graph{(y^2-8y+x^2-4x+19)(y^2-6y+x^2-10x+30)=0 [-14.24, 14.24, -7.11, 7.13]}