Circle A has a center at #(1 ,4 )# and a radius of #5 #. Circle B has a center at #(9 ,3 )# 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 #(1   ,4  )# and a radius of #5 #.  Circle B has a center at #(9   ,3  )# and a radius of #1 #.  Do the circles overlap?  If not what is the smallest distance between them?

Evelyn Arboleda · Accepted Answer

There is a minimum distance of #(sqrt(65)-6)~~2.06# units between the two (non-overlapping) circles.

The distance between the two centers is #color(white)("XXX")d=sqrt((9-1)^2+(4-3)^2)=sqrt(65)~~8.06# Along the line segment connecting the two centers #5# units are covered by circle A, and #1# unit is covered by circle B. So only #5+1=6# units are covered by the circles. Leaving #sqrt(65)-6~~2.06# units uncovered.

Theodore Abad · Answer

no overlap , ≈ 2.06

What we have to do here is compare the distance (d) 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 To calculate the distance between the centres use the #color(blue)" distance formula "# #color(red)(|bar(ul(color(white)(a/a)color(black)(d=sqrt((x_2-x_1)^2+(y_2-y_1)^2))color(white)(a/a)|)))# where # (x_1,y_1)" and " (x_2,y_2)" are 2 points"# let # (x_1,y_1)=(1,4)" and " (x_2,y_2)=(9,3)# #d=sqrt((9-1)^2+(3-4)^2)=sqrt65 ≈ 8.06# radius of A + radius of B = 5 + 1 = 6 Since sum of radii < d , then no overlap smallest distance = 8.06 - 6 = 2.06 graph{(y^2-8y+x^2-2x-8)(y^2-6y+x^2-18x+89)=0 [-35.56, 35.56, -17.78, 17.78]}

James Atkins · Answer

undefined The circles do not overlap. The smallest distance between them is the distance between their centers minus the sum of their radii.

First, find the distance between the centers using the distance formula:

$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $

$ d = \sqrt{(9 - 1)^2 + (3 - 4)^2} $

$ d = \sqrt{64 + 1} $

$ d = \sqrt{65} $

Now, subtract the sum of their radii from this distance:

$ \text{Smallest distance} = \sqrt{65} - (5 + 1) $

$ \text{Smallest distance} = \sqrt{65} - 6 $

So, the smallest distance between the circles is $ \sqrt{65} - 6 $ units.