# Circle A has a radius of 2  and a center of (3 ,7 ). Circle B has a radius of 6  and a center of (8 ,1 ). If circle B is translated by <-4 ,3 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Apr 5, 2016

The distance between the centers of the circle is less than the sum of the length of their radiuses. Thus the circles overlap and circle ${O}_{B} '$ totally engulf circle ${O}_{A}$

$| 3.16 | < | 2 + 6 |$

#### Explanation:

Given : Circles A, B with centers and radiuses
O_A(3,7;r=2); => (x-3)^2+(y-7)^2=4
O_B(8,1;r=6); => (x-8)^2+(y-1)^2=36

Required :
If circle O_B(8,1;r=6) translates by (-4,3) to =>O_(B')(x,y;r=6)
Does O_(B')(x,y;r=6) overlap with O_A(3,7;r=2)

Solution Strategy:
a) Translate ${O}_{B}$ by vector <-4,3> and find the new center
b) Calculate the distance from ${O}_{A} \left(3 , 7\right) \iff {O}_{B} ' \left(x , y\right) = \overline{A B '}$
c) Is $| \overline{A B '} | < | {r}_{A} + {r}_{B} ' |$
If the above is $\text{holds" => "it overlaps" if "False" => "it does not}$

a) Translation Matrix of center vector $< 3 , 7 >$ by <-4,3> is
${T}_{- 4 , 3} \left[\begin{matrix}8 \\ 1\end{matrix}\right] = \left[\begin{matrix}1 - 4 & 0 \\ 0 & 1 + 3\end{matrix}\right] \left[\begin{matrix}8 \\ 1\end{matrix}\right] = \left[\begin{matrix}8 - 4 \\ 1 + 3\end{matrix}\right] = \left[\begin{matrix}4 \\ 4\end{matrix}\right]$
So the new center ${O}_{B '} \left(x , y\right) = {O}_{B '} \left(4 , 4\right)$

b) Use the distance formula to calculate ${O}_{A} \left(3 , 7\right) \iff {O}_{B} ' \left(4 , 4\right)$
$\overline{A B '} = \sqrt{{\left(3 - 4\right)}^{2} + {\left(7 - 4\right)}^{2}} = \sqrt{{\left(- 1\right)}^{2} + {\left(3\right)}^{2}} = \sqrt{10} = 3.16$

c) Is Is $| \overline{A B '} | < | {r}_{A} + {r}_{B} ' |$
$| 3.16 | < | 2 + 6 |$
Since the distance from the center is less that the radius of ${O}_{B} '$, (about half of ${O}_{B} '$), the circles not only do they overlap but also ${O}_{B} '$ totally engulf ${O}_{A}$