Circle A has a center at #(3 ,2 )# and an area of #13 pi#. Circle B has a center at #(9 ,6 )# and an area of #28 pi#. Do the circles overlap?

Question

Circle A has a center at #(3   ,2  )# and an area of #13 pi#.  Circle B has a center at #(9  ,6   )# and an area of #28 pi#.  Do the circles overlap?

Sadie Adams · Accepted Answer

Circles Overlap

In Circle A #pi R_A^2 = A_A = 13pi#

#R_A = sqrt((cancelpi * 13) / cancel(pi)) = sqrt13 #

Similarly in Circle B, #R_B = sqrt((cancel pi * 28)/cancel(pi)) = sqrt28 #

#R_A + R_B = sqrt13 + sqrt28 ~~ 8.8971#

Distance between the centers #O_A, O_B#

#vec(O_AO_B) = sqrt(9-6)^2 + (6-2)^2) = sqrt52 ~~ 7.2111#

Since #vec(O_AO_B) < R_A + R_B#, both the circles overlap.

Noah Atkinson · Answer

undefined To determine if the circles overlap, we need to compare the distance between their centers to the sum of their radii. If the distance between the centers is less than the sum of their radii, the circles overlap.

Let's denote the centers of the circles as $ (x_1, y_1) $ for Circle A and $ (x_2, y_2) $ for Circle B. The formula for the distance between two points $ (x_1, y_1) $ and $ (x_2, y_2) $ is given by:

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

For Circle A with center $ (3, 2) $ and Circle B with center $ (9, 6) $, the distance between their centers is:

$$ \sqrt{(9 - 3)^2 + (6 - 2)^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} $$

Now, let's find the radii of Circle A and Circle B. The area of a circle is given by $ \pi r^2 $, so we can solve for the radii:

For Circle A: 
$$ 13\pi = \pi r^2 $$
$$ r^2 = \frac{13\pi}{\pi} = 13 $$
$$ r_A = \sqrt{13} $$

For Circle B: 
$$ 28\pi = \pi r^2 $$
$$ r^2 = \frac{28\pi}{\pi} = 28 $$
$$ r_B = \sqrt{28} $$

Now, we compare the distance between the centers ($ \sqrt{52} $) to the sum of their radii ($ \sqrt{13} + \sqrt{28} $):

$$ \sqrt{52} \stackrel{?}{<} \sqrt{13} + \sqrt{28} $$

$$ \sqrt{52} \stackrel{?}{<} \sqrt{13} + \sqrt{28} $$
$$ \sqrt{52} \stackrel{?}{<} \sqrt{13} + \sqrt{28} $$
$$ \sqrt{52} \approx 7.21 $$
$$ \sqrt{13} + \sqrt{28} \approx 7.21 $$

Since the distance between the centers ($ \sqrt{52} $) is equal to the sum of their radii ($ \sqrt{13} + \sqrt{28} $), the circles only touch each other externally. They do not overlap.