Circle A has a center at #(2 ,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 #(2   ,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?

Ella Armstrong · Accepted Answer

#"no overlap ",~~1.07# What we have to do here is #color(blue)"compare"# the distance (d) between the centres of the circles to the #color(blue)"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)(2/2)color(black)(d=sqrt((x_2-x_1)^2+(y_2-y_1)^2))color(white)(2/2)|)))# #"let "(x_1,y_1)=(2,4)" and "(x_2,y_2)=(9,3)# #d=sqrt((9-2)^2+(3-4)^2)=sqrt(49+1)=sqrt50~~7.07# #"sum of radii "=5+1=6# #"since sum of radii"< d" then no overlap"# #"smallest distance "=d-" sum of radii"# #color(white)(xxxxxxxxxxxx)=7.07-6=1.07# graph{((x-2)^2+(y-4)^2-25)((x-9)^2+(y-3)^2-1)=0 [-20, 20, -10, 10]}

Sadie Adams · Answer

undefined To determine if the circles overlap, we need to calculate the distance between their centers and compare it to the sum of their radii. If the distance between the centers is greater than the sum of the radii, then the circles do not overlap. Otherwise, they overlap.

Let's calculate the distance between the centers of Circle A and Circle B using the distance formula:

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

For Circle A with center $ (2, 4) $ and Circle B with center $ (9, 3) $, the distance is:

$$ \text{Distance} = \sqrt{(9 - 2)^2 + (3 - 4)^2} $$
$$ \text{Distance} = \sqrt{(7)^2 + (-1)^2} $$
$$ \text{Distance} = \sqrt{49 + 1} $$
$$ \text{Distance} = \sqrt{50} $$

Now, compare the distance to the sum of the radii:

$$ \text{Sum of radii} = \text{radius of Circle A} + \text{radius of Circle B} $$
$$ \text{Sum of radii} = 5 + 1 = 6 $$

Since $ \sqrt{50} > 6 $, the circles do not overlap.

To find the smallest distance between them, subtract the sum of the radii from the distance between their centers:

$$ \text{Smallest distance} = \sqrt{50} - 6 $$

$$ \text{Smallest distance} = \sqrt{50} - 6 \approx 0.35 $$

So, the smallest distance between the circles is approximately $ 0.35 $ units.