Circle A has a center at #(1 ,8 )# and an area of #32 pi#. Circle B has a center at #(2 ,3 )# and an area of #28 pi#. Do the circles overlap?

Question

Circle A has a center at #(1   ,8  )# and an area of #32 pi#.  Circle B has a center at #(2  ,3   )# and an area of #28 pi#.  Do the circles overlap?

Claire Akin · Accepted Answer

The circles overlap.

Distance from the center of Circle A to center of Circle B: #color(white)("XXX")d=sqrt((2-1)^2+(3-8)^2)# #color(white)("XXXX")=sqrt(26)~~5.1# Area of Circle (in general) #=pir^2# #"Area"_A = pi*r_A^2 = 32pi# #color(white)("XXX")r_A=sqrt(32)=4sqrt(2)~~5.7# Since the radius of Circle A is greater than the distance between the centers of the two circles, the center of Circle B is inside Circle A and the circles must overlap

Caroline Aucoin · 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 the radii, the circles overlap. If the distance between the centers is equal to the sum of the radii, the circles touch at one point. Otherwise, if the distance between the centers is greater than the sum of the radii, the circles do not overlap.

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

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

For Circle A with center $(1, 8)$ and Circle B with center $(2, 3)$:

$$ \text{Distance} = \sqrt{(2 - 1)^2 + (3 - 8)^2} $$
$$ = \sqrt{1^2 + (-5)^2} $$
$$ = \sqrt{1 + 25} $$
$$ = \sqrt{26} $$

The radius of Circle A (denoted as $r_A$) is calculated from its area:

$$ \text{Area}_A = \pi r_A^2 $$
$$ 32\pi = \pi r_A^2 $$
$$ r_A = \sqrt{32} = 4\sqrt{2} $$

The radius of Circle B (denoted as $r_B$) is calculated similarly:

$$ \text{Area}_B = \pi r_B^2 $$
$$ 28\pi = \pi r_B^2 $$
$$ r_B = \sqrt{28} = 2\sqrt{7} $$

The sum of the radii ($r_A + r_B$) is:

$$ 4\sqrt{2} + 2\sqrt{7} $$

Comparing the distance between the centers ($\sqrt{26}$) to the sum of the radii, if $\sqrt{26}$ is less than $4\sqrt{2} + 2\sqrt{7}$, the circles overlap. Otherwise, they do not overlap.

Since $\sqrt{26} < 4\sqrt{2} + 2\sqrt{7}$, the circles overlap.