Circle A has a center at #(3 ,7 )# and an area of #56 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   ,7  )# and an area of #56 pi#.  Circle B has a center at #(9  ,6   )# and an area of #28 pi#.  Do the circles overlap?

James Atkins · Accepted Answer

the circles will overlap.

the circles will overlap if the distance between the centers is smaller than the sum of the radius. the distance between the radius is, #d=sqrt((9-3)^2+(7-6)^2)# #=sqrt(6^2+1^2)# #=sqrt(36+1)# #=6.08# here, #pir_1^2=56pi# #or,r_1^2=56# #or,r_1=7.48# again, #pir_2^2=28pi# #or,r_2^2=28# #or,r_2=5.29# so, #r_1+r_2=7.48+5.29# #=12.77# so, the circles will overlap.

Aurora Ayala · 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.

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

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

For Circle A with center $(3, 7)$ and Circle B with center $(9, 6)$:

$$d = \sqrt{(9 - 3)^2 + (6 - 7)^2} = \sqrt{36 + 1} = \sqrt{37}$$

Now, let's calculate the sum of their radii:

- Circle A has an area of $56\pi$, so its radius $r_A$ can be found using the formula for the area of a circle:

$$A_A = \pi r_A^2$$
$$56\pi = \pi r_A^2$$
$$r_A^2 = 56$$
$$r_A = \sqrt{56}$$

- Circle B has an area of $28\pi$, so its radius $r_B$ can be found similarly:

$$A_B = \pi r_B^2$$
$$28\pi = \pi r_B^2$$
$$r_B^2 = 28$$
$$r_B = \sqrt{28}$$

The sum of the radii, $r_A + r_B$, is:

$$\sqrt{56} + \sqrt{28}$$

Since $r_A + r_B > \sqrt{37}$, the circles do not overlap.