# Circle A has a radius of 2  and a center of (6 ,3 ). Circle B has a radius of 3  and a center of (2 ,4 ). If circle B is translated by <1 ,3 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Dec 3, 2016

circles have one point of contact.

#### 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

#• If sum of radii = d , then one point of contact

Before calculating d, we have 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.

$\text{Under a translation of } \left(\begin{matrix}1 \\ 3\end{matrix}\right)$

$\left(2 , 4\right) \to \left(2 + 1 , 4 + 3\right) \to \left(3 , 7\right) \leftarrow \text{ new centre of B}$

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

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

let $\left({x}_{1} , {y}_{1}\right) = \left(6 , 3\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(3 , 7\right)$

$d = \sqrt{{\left(3 - 6\right)}^{2} + {\left(7 - 3\right)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$

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

Since sum of radii = d , then circles have one point of contact.
graph{(y^2-6y+x^2-12x+41)(y^2-14y+x^2-6x+49)=0 [-25.64, 25.68, -12.82, 12.83]}