Circle A has a center at #(3 ,7 )# and a radius of #2 #. Circle B has a center at #(1 ,3 )# and a radius of #4 #. Do the circles overlap? If not, what is the smallest distance between them?

Question

Circle A has a center at #(3  ,7   )# and a radius of #2 #.  Circle B has a center at #(1   ,3    )# and a radius of #4 #.  Do the circles overlap?  If not, what is the smallest distance between them?

Claire Akin · Accepted Answer

The circles overlap.

The distance between the center of A at #(3,7)# and B at #(1,3)# is given by the Pythagorean Theorem as #color(white)("XXX")d=sqrt((3-1)^2+(7-3)^2)=sqrt(2^2+4^2)=sqrt(20)~~4.47# Circle A with a radius of #2# covers #2# units of the distance between the centers of A and B. Circle B with a radius of #4# covers #4# units of the distance between the centers of A and B. Together the circles cover #2+4=6# which is more than the distance between the centers; so the circles overlap (by approximately #6-4.47=1.53# units).

Autumn Armstrong · Answer

undefined The distance between the centers of two circles can be calculated using the distance formula, which is the square root of the sum of the squares of the differences in their x-coordinates and y-coordinates.

For Circle A with center at (3, 7) and Circle B with center at (1, 3), the distance between their centers is:

$$ \sqrt{(3 - 1)^2 + (7 - 3)^2} $$

$$ = \sqrt{2^2 + 4^2} $$

$$ = \sqrt{4 + 16} $$

$$ = \sqrt{20} $$

$$ = 2\sqrt{5} $$

Since the distance between the centers of the two circles (2√5) is greater than the sum of their radii (2 + 4 = 6), the circles do not overlap. The smallest distance between them is the difference between the distance of their centers and the sum of their radii, which is $2\sqrt{5} - 6$.