Circle A has a center at #(2 ,2 )# and an area of #8 pi#. Circle B has a center at #(13 ,6 )# and an area of #54 pi#. Do the circles overlap?

Question

Circle A has a center at #(2   ,2  )# and an area of #8 pi#.  Circle B has a center at #(13  ,6 )# and an area of #54 pi#.  Do the circles overlap?

Savannah Aguilar · Accepted Answer

No, they don't overlap.

I start calculating the radius of the two circles. We know that #A=pir^2# then #r=sqrt(A/pi)# The radius of the circles, that I will call respectively #r_1# and #r_2# are #r_1=sqrt(8pi/pi)=sqrt(8)\approx2.83# #r_2=sqrt(54pi/pi)=sqrt(54)\approx7.35#. If the distance between the two centers is bigger than the sum of the two radius, the two circles will not overlap, otherwise they will overlap. The distance between the two centers is #d=sqrt((2-13)^2+(2-6)^2)=sqrt((-11)^2+(-4)^2)# #=sqrt(137)\approx11.7#. The sum of the two radius is #2.83+7.35=10.18# and #11.7> 10.18# so the circles cannot overlap.

Maverick Smith · Answer

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

Let's denote the center of Circle A as $ (x_1, y_1) $ and the center of Circle B as $ (x_2, y_2) $. The distance between the centers can be found using the distance formula:

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

Given the centers:
Center of Circle A: $ (2, 2) $
Center of Circle B: $ (13, 6) $

$$ d = \sqrt{(13 - 2)^2 + (6 - 2)^2} $$
$$ d = \sqrt{(11)^2 + (4)^2} $$
$$ d = \sqrt{121 + 16} $$
$$ d = \sqrt{137} $$

Now, let's find the radii of each circle. The radius $ r_1 $ of Circle A can be calculated using its area $ A_1 $:

$$ A_1 = \pi r_1^2 $$
$$ 8\pi = \pi r_1^2 $$
$$ r_1^2 = \frac{8\pi}{\pi} $$
$$ r_1^2 = 8 $$
$$ r_1 = \sqrt{8} = 2\sqrt{2} $$

Similarly, the radius $ r_2 $ of Circle B can be calculated using its area $ A_2 $:

$$ A_2 = \pi r_2^2 $$
$$ 54\pi = \pi r_2^2 $$
$$ r_2^2 = \frac{54\pi}{\pi} $$
$$ r_2^2 = 54 $$
$$ r_2 = \sqrt{54} = 3\sqrt{6} $$

Now, let's compare the sum of the radii to the distance between the centers:

$$ r_1 + r_2 = 2\sqrt{2} + 3\sqrt{6} $$

Since $ \sqrt{137} > 2\sqrt{2} + 3\sqrt{6} $, the circles do not overlap.