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?

1 Answer
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|

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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(3,7) <=> O_B'(x,y)=bar(AB')
c) Is |bar(AB')|<|r_A+r_B'|
If the above is "holds" => "it overlaps" if "False" => "it does not"

a) Translation Matrix of center vector <3,7> by <-4,3> is
T_(-4,3)[(8),(1)]= [(1-4,0),(0,1+3)] [(8),(1)]= [(8-4 ),(1+3)] =[(4),(4)]
So the new center O_(B')(x,y)=O_(B')(4,4)

b) Use the distance formula to calculate O_A(3,7) <=> O_B'(4,4)
bar(AB') = sqrt((3-4)^2+(7-4)^2) = sqrt((-1)^2+(3)^2) = sqrt(10)=3.16

c) Is Is |bar(AB')|<|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