Circle A has a center at #(8 ,4 )# and a radius of #3 #. Circle B has a center at #(-2 ,-2 )# 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 #(8   ,4   )# and a radius of #3 #.  Circle B has a center at #(-2   ,-2    )# and a radius of #4 #.  Do the circles overlap?  If not, what is the smallest distance between them?

Allison Allen · Accepted Answer

no overlap , ≈ 4.662

What we have to do here is to compare the distance (d ) between the centres with the sum of the radii. • If sum of radii > d , then circles overlap • If sum of radii < d , then no overlap To calculate d , 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)=(8,4)" and " (x_2,y_2)=(-2,-2)# #d=sqrt((-2-8)^2+(-2-4)^2)=sqrt136≈11.662# radius of A + radius of B = 3 + 4 = 7 Since sum of radii < d , then no overlap and smallest distance between them = 11.662 - 7 = 4.662 graph{(y^2-8y+x^2-16x+71)(y^2+4y+x^2+4x-8)=0 [-20, 20, -10, 10]}

Claire Akin · Answer

undefined The circles do overlap. The distance between their centers is greater than the sum of their radii. Therefore, the circles intersect. The smallest distance between the circles occurs at the points where the line connecting their centers intersects the circles. To find this distance, you can calculate the distance between the centers of the circles and then subtract the sum of their radii.

The distance between the centers of the circles can be found using the distance formula, which is the square root of the sum of the squares of the differences in their coordinates:

$$ \sqrt{(8 - (-2))^2 + (4 - (-2))^2} = \sqrt{10^2 + 6^2} = \sqrt{100 + 36} = \sqrt{136} $$

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

$$ \sqrt{136} - (3 + 4) = \sqrt{136} - 7 $$

So, the smallest distance between the circles is $ \sqrt{136} - 7 $.