# Circle A has a radius of 2  and a center of (6 ,5 ). Circle B has a radius of 3  and a center of (2 ,4 ). If circle B is translated by <1 ,1 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

May 14, 2018

$\text{circles overlap}$

#### Explanation:

$\text{what we have to do here is compare the distance (d)}$
$\text{between the centres to the sum of the radii}$

• " if sum of radii">d" then circles overlap"

• " if sum of radii"< d" then no overlap"

$\text{before calculating d we require to find the new centre}$
$\text{of B after the given translation}$

$\text{under the translation } < 1 , 1 >$

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

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

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$\text{let "(x_1,y_1)=(6,5)" and } \left({x}_{2} , {y}_{2}\right) = \left(3 , 5\right)$

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

$\text{sum of radii } = 2 + 3 = 5$

$\text{since sum of radii">d" then circles overlap}$
graph{((x-6)^2+(y-5)^2-4)((x-3)^2+(y-5)^2-9)=0 [-20, 20, -10, 10]}

May 14, 2018

The distance between the centers is $3$, which satisfies the triangle inequality with the two radii of $2$ and $3$, so we have overlapping circles.

#### Explanation:

I thought I did this one already.

A is $\left(6 , 5\right)$ radius $2$

B's new center is $\left(2 , 4\right) + < 1 , 1 > = \left(3 , 5\right) ,$ radius still $3$

Distance between centers,

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

Since the distance between the centers is less than the sum of the two radii, we have overlapping circles.