Circle A has a center at #(7 ,5 )# and an area of #92 pi#. Circle B has a center at #(9 ,2 )# and an area of #14 pi#. Do the circles overlap?

Question

Circle A has a center at #(7   ,5  )# and an area of #92 pi#.  Circle B has a center at #(9  ,2   )# and an area of #14 pi#.  Do the circles overlap?

Evelyn Arboleda · Accepted Answer

They will overlap,practically smaller one will be inside larger

Given
Area of 1st circle #=92pi=pir_1^2# , where #r_1#=radius of 1st circle
Hence #r_1=sqrt92#
Again Area of 2nd circle #=14pi=pir_2^2# , where #r_2#=radius of 2nd circle
Hence #r_1=sqrt14#
Sum of their radii #S=r_1+r_2=sqrt92+sqrt14#

Now the distance between their centers
#d=sqrt((7-9)^2+(5-2)^2)=sqrt13=#
S>d => They will overlap,practically smaller one will be inside larger one because # r_2-r_1>d#

Avery Allen · Answer

undefined To determine if the circles overlap, we need to check if the distance between their centers is less than the sum of their radii. The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

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

For Circle A with center $(7, 5)$ and Circle B with center $(9, 2)$, the distance between their centers is:

$$d = \sqrt{(9 - 7)^2 + (2 - 5)^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}$$

The radius of Circle A can be found using its area:

$$A = \pi r^2$$
$$92\pi = \pi r^2$$
$$r^2 = 92$$
$$r = \sqrt{92} = 2\sqrt{23}$$

The radius of Circle B can be found similarly:

$$14\pi = \pi r^2$$
$$r^2 = 14$$
$$r = \sqrt{14}$$

Since the sum of the radii is $2\sqrt{23} + \sqrt{14}$, we need to compare this to the distance between the centers, $\sqrt{13}$, to determine if the circles overlap. If the distance is less than the sum of the radii, the circles overlap. Otherwise, they do not overlap.