# Circle A has a radius of 4  and a center of (7 ,3 ). Circle B has a radius of 2  and a center of (1 ,2 ). 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?

Jun 7, 2016

circles overlap

#### Explanation:

What we have to do here is compare the distance (d ) between the centres with the sum and difference of the radii.

• If sum of radii > d , then circles overlap

• If sum of radii < d , then no overlap

The first step is to calculate the new centre of B under the given translation. Note that a translation does not change the shape of the figure , only it's position.

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

B (1 ,2) → (1+2 ,2+4) → (3 ,6)

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

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

The 2 points here are (3 ,6) and (7 ,3)

$d = \sqrt{{\left(7 - 3\right)}^{2} + {\left(3 - 6\right)}^{2}} = \sqrt{25} = 5$

radius of A + radius of B = 4 + 2 = 6

Since sum of radii > d , then circles overlap
graph{(y^2-6y+x^2-14x+42)(y^2-12y+x^2-6x+41)=0 [-20, 20, -10, 10]}