Circle A has a center at #(5 ,4 )# and a radius of #3 #. Circle B has a center at #(6 ,-8 )# and a radius of #1 #. Do the circles overlap? If not, what is the smallest distance between them?

Question

Circle A has a center at #(5  ,4   )# and a radius of #3    #.  Circle B has a center at #(6   ,-8   )# and a radius of #1 #.  Do the circles overlap?  If not, what is the smallest distance between them?

Violet Aldrich · Accepted Answer

no overlap ,d ≈ 8.042 units

First step is to calculate the distance between the centres using the #color(blue)" distance formula "# #d =sqrt((x_2-x_1)^2 + (y_2-y_1)^2 # where #(x_1,y_1)" and "(x_2,y_2)" are 2 coordinate points "# let #(x_1,y_1)=(5,4)" and " (x_2,y_2)=(6,-8)# #rArr d = sqrt((6-5)^2 +(-8-4)^2) = sqrt145 ≈ 12.042# now: radius of A + radius of B = 3 + 1 = 4 since 4 < 12.042 the circles do not overlap and distance between them = 12.042 - 4 = 8.042

Sadie Adams · Answer

undefined The circles do not overlap. The smallest distance between them is the distance between their centers minus the sum of their radii. This distance is calculated as follows:

$$ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} - (r_1 + r_2) $$

For Circle A with center $(5, 4)$ and radius $3$, and Circle B with center $(6, -8)$ and radius $1$, the distance between their centers is:

$$ \text{Distance} = \sqrt{(6 - 5)^2 + (-8 - 4)^2} - (3 + 1) $$
$$ \text{Distance} = \sqrt{1^2 + (-12)^2} - 4 $$
$$ \text{Distance} = \sqrt{1 + 144} - 4 $$
$$ \text{Distance} = \sqrt{145} - 4 $$

So, the smallest distance between the circles is $ \sqrt{145} - 4 $, approximately equal to $9.54 - 4 = 5.54$.