Circle A has a center at #(12 ,9 )# and an area of #13 pi#. Circle B has a center at #(3 ,1 )# and an area of #28 pi#. Do the circles overlap?

Question

Circle A has a center at #(12   ,9  )# and an area of #13 pi#.  Circle B has a center at #(3  ,1   )# and an area of #28 pi#.  Do the circles overlap?

Sadie Allen · Accepted Answer

No

Circle #A# has a centre #(12,9)# and an area of #13 pi#. Thus:

# pi(r_A)^2 = 13pi => r_A=sqrt(13) ~~ 3.61 #

Circle #B# has a centre #(3,1 )# and an area of #28 pi#. Thus:

# pi(r_B)^2 = 28pi => r_B=sqrt(28) ~~ 5.29 #

We can find the distance #AB# using Pythagoras:

# AB^2 = (12-3)^2 + (9-1)^2 #
# " " = 9^2 + 8^2 #
# " " = 81+64 #
# " " = 145 => AB = sqrt(145) ~~ 12.04#

And as #AB gt r_A _ r_B => # The circles do not overlap.

Carson Arnold · Answer

undefined To determine if the circles overlap, we can check if the distance between their centers is less than the sum of their radii. If it is, then the circles overlap.

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

Distance $ d $ between two points $ (x_1, y_1) $ and $ (x_2, y_2) $ is given by:
$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

For Circle A with center $ (12, 9) $ and Circle B with center $ (3, 1) $:
$$ d = \sqrt{(3 - 12)^2 + (1 - 9)^2} = \sqrt{(-9)^2 + (-8)^2} = \sqrt{81 + 64} = \sqrt{145} $$

Now, we find the radii of the circles. The radius $ r_1 $ of Circle A is:
$$ \text{Area of Circle A} = 13\pi = \pi r_1^2 $$
$$ r_1^2 = 13 $$
$$ r_1 = \sqrt{13} $$

The radius $ r_2 $ of Circle B is:
$$ \text{Area of Circle B} = 28\pi = \pi r_2^2 $$
$$ r_2^2 = 28 $$
$$ r_2 = 2\sqrt{7} $$

Now we compare the distance between the centers $ d $ to the sum of their radii $ r_1 + r_2 $:

$$ \sqrt{145} \stackrel{?}{<} \sqrt{13} + 2\sqrt{7} $$

Since $ \sqrt{145} > \sqrt{13} + 2\sqrt{7} $, the circles do not overlap.