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

Sep 12, 2016

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

• If sum of radii > d , then circles overlap

• If sum of radii < d , then no overlap

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

Under translation $\left(\begin{matrix}2 \\ 1\end{matrix}\right)$

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

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

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points}$

here the 2 points are (1 ,2) and (5 ,5) the centres of the circles.

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

$d = \sqrt{{\left(5 - 1\right)}^{2} + {\left(5 - 2\right)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$

Sum of radii = radius of A + radius of B = 2 + 5 = 7

Since sum of radii > d , then circles overlap.
graph{(y^2-4y+x^2-2x+1)(y^2-10y+x^2-10x+25)=0 [-22.5, 22.5, -11.25, 11.25]}